In mathematics, the four color theorem, or the four color map theorem, states that, given any separation of a plane into contiguous regions, producing a figure called a map, no more than four colors are required to color the regions of the map so that no two adjacent regions have the same color.
— Wikipedia, the free encyclopedia 

In this problem, you have to solve the 4-color problem. Hey, I’m just joking. 

You are asked to solve a similar problem: 

Color an N × M chessboard with K colors numbered from 1 to K such that no two adjacent cells have the same color (two cells are adjacent if they share an edge). The i-th color should be used in exactly c _icells. 

Matt hopes you can tell him a possible coloring.

InputThe first line contains only one integer T (1 ≤ T ≤ 5000), which indicates the number of test cases. 

For each test case, the first line contains three integers: N, M, K (0 < N, M ≤ 5, 0 < K ≤ N × M ). 

The second line contains K integers c _i (c _i > 0), denoting the number of cells where the i-th color should be used. 

It’s guaranteed that c ₁ + c ₂ + · · · + c _K = N × M . 
OutputFor each test case, the first line contains “Case #x:”, where x is the case number (starting from 1). 

In the second line, output “NO” if there is no coloring satisfying the requirements. Otherwise, output “YES” in one line. Each of the following N lines contains M numbers seperated by single whitespace, denoting the color of the cells. 

If there are multiple solutions, output any of them.Sample Input

4
1 5 2
4 1
3 3 4
1 2 2 4
2 3 3
2 2 2
3 2 3
2 2 2

Sample Output

Case #1:
NO
Case #2:
YES
4 3 4
2 1 2
4 3 4
Case #3:
YES
1 2 3
2 3 1
Case #4:
YES
1 2
2 3
3 1
没什么好说的，dfsdfsdfs

//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include<cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long LL;
const LL INF = 1e13;
const int mod = 1000000007;
const int mx = 1e7; //check the limits, dummy
typedef pair<int, int> pa;
//const double PI = acos(-1);
//ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
#define swa(a,b) a^=b^=a^=b
#define re(i,a,b) for(int i=(a),_=(b);i<_;i++)
#define rb(i,a,b) for(int i=(b),_=(a);i>=_;i--)
#define clr(a) memset(a, 0, sizeof(a))
#define lowbit(x) ((x)&(x-1))
//#define mkp make_pai
void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); }
int n, m, k,ans;
int a[mx],mp[55][55];
bool flag;
bool dfs(int x, int y, int r) {
    if (!r) {
        flag = 1;
        return 1;
    }
    re(i, 1, k + 1)if (a[i] > (r + 1) / 2)return 0;
    re(i, 1, k + 1) {
        if (a[i]) {
            if (x && mp[x - 1][y] == i)continue;
            if (y && mp[x][y - 1] == i)continue;
            a[i]--;
            mp[x][y] = i;
            if (y < m - 1)dfs(x, y + 1, r - 1);
            else dfs(x + 1, 0, r - 1);
            if (flag)return 1;
            a[i]++;
        }
    }
    return 0;
}
int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T, cas = 1;
    scanf("%d", &T);
    while (T--)
    {
        clr(mp);
        flag = 0;
        sc(n), sc(m), sc(k);
        re(i, 1, k+1)sc(a[i]);

        printf("Case #%d:\n",cas++);
        if (!dfs(0, 0, n * m)) {
            puts("NO");
            continue;
        }
        puts("YES");
        re(i, 0, n)re(j, 0, m)printf(j == m - 1 ? "%d\n" : "%d ", mp[i][j]);
    }
    return 0;
}