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 cas, n, k, remaind;
int main() {
  scanf("%d", &cas);
  while (cas--) {
    scanf("%d", &n);
    scanf("%d", &k);
    remaind = n - k + 1;
    if (remaind > 0 && (remaind & 1)) {
      printf("YES\n");
      for (int i = 0; i < k - 1; ++i) printf("1 ");
      printf("%d\n", n - k + 1);
      continue;
    }
    remaind = n - 2 * k + 2;
    if (remaind > 0 && (remaind & 1) == 0) {
      printf("YES\n");
      for (int i = 0; i < k - 1; ++i) printf("2 ");
      printf("%d\n", n - (k << 1) + 2);
      continue;
    }
    printf("NO\n");
  }
  return 0;
}