Problem: You are given two positive integers n (1 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).

In other words, find a_1, a_2, …, a_k such that all a_i>0, n = a_1 + a_2 + … + a_k and either all a_i are even or all a_i are odd at the same time.

If such a representation does not exist, then report it.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.

Each test case is two positive integers n (1 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 100).

Output

For each test case print:

  * YES and the required values a_i, if the answer exists (if there are several answers, print any of them); 
  * NO if the answer does not exist. 



The letters in the words YES and NO can be printed in any case.

Example

Input


8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9


Output


YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int tc;
  scanf("%d", &tc);
  for (int i = 0; i < tc; i++) {
    int N, K;
    scanf("%d %d", &N, &K);
    if (N < K) {
      printf("NO\n");
      continue;
    }
    if (N % 2 == 0) {
      if (2 * K <= N) {
        printf("YES\n");
        for (int j = 0; j < K - 1; j++) {
          if (j != 0) printf(" ");
          printf("%d", 2);
        }
        printf(" %d\n", N - (2 * (K - 1)));
      } else if (N == K) {
        printf("YES\n");
        for (int j = 0; j < K; j++) {
          if (j != 0) printf(" ");
          printf("%d", 1);
        }
        printf("\n");
      } else if (K % 2 != 0) {
        printf("NO\n");
      } else {
        bool flag = false;
        int count = 0;
        while (N > K) {
          count++;
          N -= 3;
          K -= 1;
          if (N == K) {
            flag = true;
            printf("YES\n");
            for (int l = 0; l < count; l++) {
              if (l != 0) printf(" ");
              printf("%d", 3);
            }
            for (int l = 0; l < N; l++) {
              printf(" %d", 1);
            }
            printf("\n");
            break;
          }
        }
        if (!flag) {
          printf("NO\n");
        }
      }
    } else {
      if (K % 2 != 1) {
        printf("NO\n");
      } else {
        printf("YES\n");
        for (int j = 0; j < K - 1; j++) {
          if (j != 0) printf(" ");
          printf("%d", 1);
        }
        printf(" %d\n", N - ((K - 1)));
      }
    }
  }
  return 0;
}