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>
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
inline long long readll() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
using namespace std;
int main() {
  int T = read();
  while (T--) {
    int n = read(), k = read();
    if (k == 1) {
      printf("YES\n%d\n", n);
      continue;
    }
    if (n < k) {
      puts("NO");
      continue;
    }
    if (k % 2) {
      if (n % 2) {
        int t = (n / k);
        if (t % 2 == 0) --t;
        puts("YES");
        for (int i = 1; i < k; i++) printf("%d ", t);
        printf("%d\n", n - (k - 1) * t);
      } else {
        if (n < 2 * k) {
          puts("NO");
          continue;
        }
        int t = (n / k) >> 1;
        t <<= 1;
        puts("YES");
        for (int i = 1; i < k; i++) printf("%d ", t);
        printf("%d\n", n - (k - 1) * t);
      }
    } else {
      if (n % 2) {
        puts("NO");
        continue;
      }
      puts("YES");
      for (int i = 1; i < k; i++) printf("%d ", n / k);
      printf("%d\n", n - ((n / k) * (k - 1)));
    }
  }
  return 0;
}