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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int T;
  cin >> T;
  while (T--) {
    long long int t, n, i, j, x, y, m, l, d, k;
    cin >> n >> k;
    if (k > n || (n % 2 == 1 && k % 2 == 0)) {
      cout << "NO"
           << "\n";
      continue;
    }
    if (k == 1) {
      cout << "YES\n" << n << "\n";
      continue;
    }
    if (n % 2 == 1 || k % 2 == 0) {
      cout << "YES\n";
      cout << n - k + 1 << ' ';
      for (int i = 0; i < (k - 1); ++i) cout << 1 << ' ';
      cout << '\n';
    } else {
      if (2 * k > n) {
        cout << "NO"
             << "\n";
        continue;
      } else {
        cout << "YES\n";
        cout << n - 2 * (k - 1) << ' ';
        for (int i = 0; i < (k - 1); ++i) cout << 2 << ' ';
        cout << '\n';
      }
    }
  }
}