Wrong Addition

Tanya is learning how to add numbers, but so far she is not doing it correctly. She is adding two numbers aa and bb using the following algorithm:

1. If one of the numbers is shorter than the other, Tanya adds leading zeros so that the numbers are the same length.
2. The numbers are processed from right to left (that is, from the least significant digits to the most significant).
3. In the first step, she adds the last digit of aa to the last digit of bb and writes their sum in the answer.
4. At each next step, she performs the same operation on each pair of digits in the same place and writes the result to the left side of the answer.

For example, the numbers a = 17236a=17236 and b = 3465b=3465 Tanya adds up as follows:

\large{ \begin{array}{r} + \begin{array}{r} 17236\\ 03465\\ \end{array} \\ \hline \begin{array}{r} 1106911 \end{array} \end{array}}[phone]​1106911​​​

• calculates the sum of 6 + 5 = 116+5=11 and writes 1111 in the answer.
• calculates the sum of 3 + 6 = 93+6=9 and writes the result to the left side of the answer to get 911911.
• calculates the sum of 2 + 4 = 62+4=6 and writes the result to the left side of the answer to get 69116911.
• calculates the sum of 7 + 3 = 107+3=10, and writes the result to the left side of the answer to get 106911106911.
• calculates the sum of 1 + 0 = 11+0=1 and writes the result to the left side of the answer and get 11069111106911.

As a result, she gets 11069111106911.

You are given two positive integers aa and ss. Find the number bb such that by adding aa and bb as described above, Tanya will get ss. Or determine that no suitable bb exists.

Input

The first line of input data contains an integer tt (1 \le t \le 10^41≤t≤104) — the number of test cases.

Each test case consists of a single line containing two positive integers aa and ss (1 \le a \lt s \le 10^{18}1≤a<s≤1018) separated by a space.

Output

For each test case print the answer on a separate line.

If the solution exists, print a single positive integer bb. The answer must be written without leading zeros. If multiple answers exist, print any of them.

If no suitable number bb exists, output -1.

Sample 1

Inputcopy	Outputcopy
6 17236 1106911 1 5 108 112 12345 1023412 1 11 1 20	3465 4 -1 90007 10 -1

Note

The first test case is explained in the main part of the statement.

In the third test case, we cannot choose bb that satisfies the problem statement.

Brainless translation ： 

Tanya Learning how to add numbers , But so far , She didn't do it right . She is adding two numbers, one One and bb Use the following algorithm ：

1. If one number is shorter than the other ,Tanya Add leading zeros , Make the numbers the same length .
2. These numbers go from right to left （ namely , From the least significant number to the most significant number ） To deal with .
3. In the first step , She will have one One To The last digit of bb And write their total in the answer .
4. In the next step , She performs the same operation on each pair of numbers in the same position , And write the result into the answer left .

for example , Numbers a = 17236 One =17236 and b = 3465b=3465 Tanya adds up as follows ：

\large{ \begin{array}{r} + \begin{array}{r} 17236\\ 03465\\ \end{array} \\ \hline \begin{array}{r} 1106911 \end{array} \end{array}}[phone]​1106911​​​

• Calculation 6 + 5 = 116+5=11 And write 1111 In the answer .
• Calculation 3 + 6 = 103+6=9 And write the result to the left of the answer to get 911911.
• Calculation 2 + 4 = 62+4=6 And write the result to the left of the answer to get 69116911.
• Calculation 7 + 3 = 107+3=10, And write the result to the left of the answer to get 106911106911.
• Calculation 1 + 0 = 11+0=1 Write the result to the left of the answer and get 11069111106911.

result , She got 11069111106911.

You will get two positive integers, one One and ss. Find number bb such , By adding a One and bb As mentioned above , Tanya will get ss. Or make sure there is no suitable bb There is .

Input

The first row of input data contains an integer tt (1 \le t \le 10^41≤t≤104） — The number of test cases .

Each test case consists of a line containing two positive integers One and ss (1 \le a \lt s \le 10^{18}1≤ One <s≤1018） Separated by spaces .

Output

For each test case , Print the answer on a separate line .

If the solution exists , Please print a single positive integer bb. The answer must be without leading zeros . If there are multiple answers , Please print any of them .

If there is no appropriate number bb There is , Output  -1.

Example 1

Input Copy	Output Copy
6 17236 1106911 1 5 108 112 12345 1023412 1 11 1 20	3465 4 -1 90007 10 -1

Be careful

The first test case is explained in the main part of the statement .

In the third test case , We can't choose bb Meet the problem statement .

There are many situations to consider in this problem , At that time, loopholes were found bit by bit

After it is made, detailed notes are added .

 

#include<bits/stdc++.h>
using namespace std;
int n,l1,l2,ans,flag;
/// Two used c++ function .
/// Insert... From the front ：s.insert (0,"111");
///int turn string：to_string(i);
string s,sum,b;

// Remove the lead 0
string fun(string s)
{
    int cnt=0;
    for(char c:s)
    {
        if(c!='0')
        {
            break;
        }
        else if(c=='0')
        {
            cnt++;
        }
    }
    return s.substr(cnt,s.size()-cnt);
}
int main()
{
    cin>>n;
    while(n--)
    {
        flag=0;
        b="";
        cin>>s>>sum;
        /// situation 1：s and sum When equal , Should return 0
        if(s==sum)
        {
            cout<<"0"<<endl;
            continue;
        }
        l1=s.size()-1,l2=sum.size()-1;
        while(l2>=0)/// as long as sum It's not over yet , Just keep cycling 
        {
            int i;
            if(l1>=0)/// situation 2： because s And what you ask b We don't know which is longer , So if s be without ,sum also , No more s Position simulation supplement 0
            {
                i=sum[l2]-'0'-(s[l1]-'0');
            }
            else/// Here is the s Analog complement 0
            {
                i=sum[l2]-'0';
            }

            if(i>=0)/// situation 3： If the sum of two numbers is greater than or equal to 0, Just add it directly to the answer string 
            {
                b.insert(0,to_string(i));
            }
            else if(i<0 && l2>=1)/// If the subtraction of two numbers is less than 0 explain sum You have to add one , That is, add the previous number as ten digits , But the premise is sum The subscript of is greater than or equal to 1
            {
                l2--;
                i+=(sum[l2]-'0')*10;// Take the new one as ten digits 
                if(i>=10)/// If you get b It's two digits , Directly declare failure , Because according to the rules ,s and b It's a bit by bit addition 
                {
                    flag=-1;
                    break;
                }
                else if(i>=0 &&i<10)/// If you get b It's a single digit , Then the conditions are met , Add it to the answer string 
                {
                    b.insert(0,to_string(i));
                }
                else if(i<0)/// If sum Got into a get b Less than 0, It's impossible , Then declare failure .
                {
                    flag=-1;
                    break;
                }

            }
            else if(i<0 && l2<1)/// If sum Not enough to carry , Explain that it is not possible to meet , Declare failure directly 
            {
                flag=-1;
                break;
            }
            l1--,l2--;
        }

        if(l1>=0 ||l2>=0 || flag==-1)// Failure output -1
        {
            cout<<"-1"<<endl;
        }
        else// Successfully output without leading 0 Answer string b
        {
            cout<<fun(b)<<endl;
        }
    }
    return 0;
}

Vigilance ！！！！！！！！！！！！！！！！！！

Not every time c++ When you do the problem , If you need anything, just check the function and use it

Learn from every one you meet , Remember , Don't check it every time , Can't remember them .

You'll never learn like this .

Often look at what I've done before c++ Method ！！