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;
const long long LINF = 1E18;
const int INF = 1E9;
const int MOD = 1E9 + 7;
const int nmax = 1;
int t = 1;
long long n, k;
void INPUT() { cin >> t; }
void SOLVE() {
  cin >> n >> k;
  if (k > n) {
    cout << "NO" << endl;
    return;
  }
  if (n % 2 == 0) {
    if (k % 2 == 1) {
      if (2 * k <= n) {
        cout << "YES" << endl;
        for (int i = 1; i < k; i++) cout << 2 << ' ';
        cout << n - (k - 1) * 2 << endl;
        return;
      } else {
        cout << "NO" << endl;
        return;
      }
    } else {
      cout << "YES" << endl;
      for (int i = 1; i < k; i++) cout << 1 << ' ';
      cout << n - k + 1 << endl;
      return;
    }
  } else {
    if (k % 2 == 0) {
      cout << "NO" << endl;
      return;
    } else {
      if ((n - k + 1) % 2 == 1) {
        cout << "YES" << endl;
        for (int i = 1; i < k; i++) cout << 1 << ' ';
        cout << n - k + 1 << endl;
        return;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  if (fopen("test"
            ".inp",
            "r")) {
    freopen(
        "test"
        ".inp",
        "r", stdin);
    freopen(
        "test"
        ".out",
        "w", stdout);
  }
  INPUT();
  while (t--) SOLVE();
  return 0;
}