A. Round Down the Price

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

At the store, the salespeople want to make all prices round.

In this problem, a number that is a power of 1010 is called a round number. For example, the numbers 100=1100=1, 101=10101=10, 102=100102=100 are round numbers, but 2020, 110110 and 256256 are not round numbers.

So, if an item is worth mm bourles (the value of the item is not greater than 109109), the sellers want to change its value to the nearest round number that is not greater than mm. They ask you: by how many bourles should you decrease the value of the item to make it worth exactly 10k10k bourles, where the value of kk — is the maximum possible (kk — any non-negative integer).

For example, let the item have a value of 178178-bourles. Then the new price of the item will be 100100, and the answer will be 178−100=78178−100=78.

Input

The first line of input data contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases .

Each test case is a string containing a single integer mm (1≤m≤1091≤m≤109) — the price of the item.

Output

For each test case, output on a separate line a single integer dd (0≤d<m0≤d<m) such that if you reduce the cost of the item by dd bourles, the cost of the item will be the maximal possible round number. More formally: m−d=10km−d=10k, where kk — the maximum possible non-negative integer.

Example

input

Copy

7
1
2
178
20
999999999
9000
987654321

output

Copy

0
1
78
10
899999999
8000
887654321

Note

In the example:

• 1−0=1001−0=100,
• 2−1=1002−1=100,
• 178−78=102178−78=102,
• 20−10=10120−10=101,
• 999999999−899999999=108999999999−899999999=108,
• 9000−8000=1039000−8000=103,
• 987654321−887654321=108987654321−887654321=108.

Note that in each test case, we get the maximum possible round number.

Problem solving instructions ： This problem is a math problem , Find the nearest number 10 The power of , You can multiply each time by 10, Then calculate the difference .

#include<stdio.h>
int main()
{
	long long n, t, m;
	scanf("%lld", &n);
	for (long long i = 0; i<n; i++) 
	{
		scanf("%lld", &t);
		m = 1;
		while (m <= t)
		{
			m *= 10;
		}
		m /= 10;
		printf("%lld\n", t - m);
	}
	return 0;
}