A. Color the Picture- Codeforces Round ＃810 (Div. 1)

A. Color the Picture

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A picture can be represented as an n×mn×m grid (nn rows and mm columns) so that each of the n⋅mn⋅m cells is colored with one color. You have kk pigments of different colors. You have a limited amount of each pigment, more precisely you can color at most aiai cells with the ii-th pigment.

A picture is considered beautiful if each cell has at least 33 toroidal neighbors with the same color as itself.

Two cells are considered toroidal neighbors if they toroidally share an edge. In other words, for some integers 1≤x1,x2≤n1≤x1,x2≤n and 1≤y1,y2≤m1≤y1,y2≤m, the cell in the x1x1-th row and y1y1-th column is a toroidal neighbor of the cell in the x2x2-th row and y2y2-th column if one of following two conditions holds:

• x1−x2≡±1(modn)x1−x2≡±1(modn) and y1=y2y1=y2, or
• y1−y2≡±1(modm)y1−y2≡±1(modm) and x1=x2x1=x2.

Notice that each cell has exactly 44 toroidal neighbors. For example, if n=3n=3 and m=4m=4, the toroidal neighbors of the cell (1,2)(1,2) (the cell on the first row and second column) are: (3,2)(3,2), (2,2)(2,2), (1,3)(1,3), (1,1)(1,1). They are shown in gray on the image below:

The gray cells show toroidal neighbors of (1,2)(1,2).

Is it possible to color all cells with the pigments provided and create a beautiful picture?

Input

Each test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤1041≤t≤104). The description of the test cases follows.

The first line of each test case contains three integers nn, mm, and kk (3≤n,m≤1093≤n,m≤109, 1≤k≤1051≤k≤105) — the number of rows and columns of the picture and the number of pigments.

The next line contains kk integers a1,a2,…,aka1,a2,…,ak (1≤ai≤1091≤ai≤109) — aiai is the maximum number of cells that can be colored with the ii-th pigment.

It is guaranteed that the sum of kk over all test cases does not exceed 105105.

Output

For each test case, print "Yes" (without quotes) if it is possible to color a beautiful picture. Otherwise, print "No" (without quotes).

Example

input

Copy

6
4 6 3
12 9 8
3 3 2
8 8
3 3 2
9 5
4 5 2
10 11
5 4 2
9 11
10 10 3
11 45 14

output

Copy

Yes
No
Yes
Yes
No
No

Note

In the first test case, one possible solution is as follows:

In the third test case, we can color all cells with pigment 11.==========================================================================================================================================

构造题，必然是整行整列涂，很容易发现，两行或两列抱团就一定可以，大于等于两行两列也是可以的。

分别对整行涂，整列涂分情况讨论

整列涂的时候，把每种颜色能涂的最大列数算出来，计算大于等于2的总和

如果行是偶数，2 4 6 8 10...

最后发现只要合超过n就可以，由于我们都是大于等于2的，在计算合的时候，如果等于n那肯定正好，如果结果是n+1，那么一定是出现了大于等于3的奇数，把这个给扣掉就行，n+2也是同理

如果行是奇数，如果之和为n,那么一定可以且一定包含一个大于等于3的，n+1的时候，就要分情况讨论了，n+1是偶数，如果全是2那么肯定不行了，必然有一个2被我们扣掉1成为单独的，但凡有一个大于等于3的，我们就可以扣掉1，剩下的抱团组成n。n+2是奇数，如果全是大于等于3的奇数，扣掉两个不同的1，如果有一个是大于等于3的，剩下全是2，那么显然扣掉一个2就可以，且只能扣掉一个2，而且必须得有一个大于等于3的来保证是奇数。n+3也是同理

可知奇数时必须要有一个大于等于3的来维持

# include <iostream>
# include<algorithm>
# include<cstring>
# include<vector>
# include<queue>
# define mod 1000000007

using namespace std;
typedef long long int ll;
ll a[100000+10];
ll b[100000+10];

int main()
{


    int t;

    cin>>t;

    while(t--)
    {
        ll n,m,k;


        cin>>n>>m>>k;

        int flag=0;


        for(int i=1; i<=k; i++)
        {
            cin>>a[i];


        }

        if(n==1||m==1)
        {
            cout<<"NO"<<endl;

            continue;

        }
       ll cnt2=0,cnt3=0;//,flag=0;

        for(int i=1; i<=k; i++)
        {
            b[i]=a[i]/m;

            if(b[i]>=2)
            {
                cnt2+=b[i];

                if(b[i]>=3)
                {
                    cnt3=1;

                }
            }

        }

        if(n%2==0)
        {
            if(cnt2>=n)
            {
                flag=1;
            }
        }
        else if(n%2)
        {

            if(cnt2>=n)
            {
                if(cnt3)
                    flag=1;
            }

        }

        swap(n,m);


        cnt2=0,cnt3=0;
        for(int i=1; i<=k; i++)
        {
            b[i]=a[i]/m;



            if(b[i]>=2)
            {
                cnt2+=b[i];

                if(b[i]>=3)
                {
                    cnt3=1;

                }
            }

        }

        if(n%2==0)
        {
            if(cnt2>=n)
            {
                flag=1;
            }
        }
        else if(n%2)
        {

            if(cnt2>=n)
            {
                if(cnt3)
                    flag=1;
            }
        }


        if(flag)
        {
            cout<<"YES"<<endl;
        }
        else
        {
            cout<<"NO"<<endl;
        }


    }
    return 0;
}