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 dx[8] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy[8] = {1, -1, -1, 0, 1, -1, 0, 1};
const long long mod = 1e9 + 7;
const long long N = 1e6 + 5;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    if (n - k + 1 > 0 && (n - k + 1) % 2) {
      cout << "YES"
           << "\n";
      cout << n - k + 1 << " ";
      while (--k) cout << 1 << " ";
    } else if (n - 2 * (k - 1) > 0 && (n - 2 * (k - 1)) % 2 == 0) {
      cout << "YES"
           << "\n";
      cout << n - 2 * (k - 1) << " ";
      while (--k) cout << 2 << " ";
    } else
      cout << "NO";
    cout << "\n";
  }
}