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>
int t, u;
long long int n, k;
int main() {
  scanf("%d", &t);
  while (t--) {
    u = 0;
    scanf("%lld %lld", &n, &k);
    if (n % 2 == 0) {
      n = n - 2 * (k - 1);
      if (n <= 0) {
        if (k % 2 == 0) {
          n = n + 2 * (k - 1);
          n = n - k + 1;
          if (n <= 0) {
            u = 1;
            printf("NO\n");
            continue;
          } else {
            printf("YES\n");
            printf("%lld ", n);
            while (k > 1) {
              printf("1 ");
              k--;
            }
          }
        } else {
          u = 1;
          printf("NO\n");
          continue;
        }
      } else {
        printf("YES\n");
        printf("%lld ", n);
        while (k > 1) {
          printf("2 ");
          k--;
        }
      }
    } else {
      if (k % 2 == 0) {
        u = 1;
        printf("NO\n");
        continue;
      } else {
        n = n - k + 1;
        if (n <= 0) {
          u = 1;
          printf("NO\n");
          continue;
        } else {
          printf("YES\n");
          printf("%lld ", n);
          while (k > 1) {
            printf("1 ");
            k--;
          }
        }
      }
    }
    if (u == 0) printf("\n");
  }
  return 0;
}