There are m soda and today is their birthday. The 1-st soda has prepared n cakes with size 1, 2, \dots, n. Now 1-st soda wants to divide the cakes into m parts so that the total size of each part is equal. 

Note that you cannot divide a whole cake into small pieces that is each cake must be complete in the m parts. Each cake must belong to exact one of m parts.

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first contains two integers n and m (1 \le n \le 10^5, 2 \le m \le 10), the number of cakes and the number of soda.
It is guaranteed that the total number of soda in the input doesn’t exceed 1000000. The number of test cases in the input doesn’t exceed 1000.

For each test case, output "YES" (without the quotes) if it is possible, otherwise output "NO" in the first line.

If it is possible, then output m lines denoting the m parts. The first number s_i of i-th line is the number of cakes in i-th part. Then s_inumbers follow denoting the size of cakes in i-th part. If there are multiple solutions, print any of them.

4
1 2
5 3
5 2
9 3

NO
YES
1 5
2 1 4
2 2 3
NO
YES
3 1 5 9
3 2 6 7
3 3 4 8

2015 Multi-University Training Contest 6

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <set>
#include <stack>
#include <algorithm>
#define LL long long
using namespace std;
const LL MAXN = 500000 + 10;
LL n, m;
LL ans[20][MAXN];
LL A[MAXN];
LL pos[MAXN];
LL tot, tar;
bool dfs(LL dep, LL now, LL u, LL c)
{
    if(now == 0)
    {
        LL k = 0;
        while(pos[k] != -1) k++;
        pos[k] = c;
        if(dfs(dep + 1, A[k], k + 1, c)) return true;
        pos[k] = -1;
        return false;
    }
    if(now == tar)
    {
        if(dep == tot) return true;
        if(dfs(dep, 0, 0, c + 1)) return true;
    }
    for(LL i=u;i<tot;i++)
    {
        if(pos[i] == -1 && now + A[i] <= tar)
        {
            pos[i] = c;
            if(dfs(dep + 1, now + A[i], i + 1, c)) return true;
            pos[i] = -1;
        }
    }
    return  false;
}
int main()
{
    LL T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        for(LL i=0;i<m;i++) for(LL j=0;j<n;j++) ans[i][j] = 0;
        if((n * (n + 1) / 2) % m != 0 || (2 * m - 1) > n)
        {
            printf("NO\n");
            continue;
        }
        while(n >= 40)
        {
            for(LL i=0;i<m;i++) ans[i][++ans[i][0]] = n - i;
            for(LL i=0;i<m;i++) ans[i][++ans[i][0]] = n - 2 * m + 1 + i;
            n -= 2 * m;
        }
        tot = n;
        tar = n * (n + 1) / (2 * m);
        for(LL i=0;i<tot;i++) A[i] = tot - i;
        for(LL i=0;i<tot;i++) pos[i] = -1;
        dfs(0, 0, 0, 0);
        for(LL i=0;i<tot;i++) ans[pos[i]][++ans[pos[i]][0]] = A[i];
        printf("YES\n");
        for(LL i=0;i<m;i++)
        {
            printf("%d", ans[i][0]);
            for(LL j=1;j<=ans[i][0];j++) printf(" %d", ans[i][j]);
            printf("\n");
        }
    }
    return 0;
}