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;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> ans1, ans2;
  for (long long j = 0; j + 1 < k; j++) ans1.push_back(1);
  ans1.push_back(n - k + 1);
  for (long long j = 0; j + 1 < k; j++) ans2.push_back(2);
  ans2.push_back(n - 2 * k + 2);
  long long cnt = 0;
  for (long long j = 0; j < ans1.size(); j++) {
    if (ans1[j] > 0 && ans1[j] & 1)
      cnt++;
    else
      break;
  }
  if (cnt == ans1.size()) {
    cout << "YES\n";
    for (long long j = 0; j < ans1.size(); j++) cout << ans1[j] << " ";
    cout << endl;
  } else {
    cnt = 0;
    for (long long j = 0; j < ans2.size(); j++) {
      if (ans2[j] > 0 && ans2[j] % 2 == 0)
        cnt++;
      else
        break;
    }
    if (cnt == ans2.size()) {
      cout << "YES\n";
      for (long long j = 0; j < ans2.size(); j++) cout << ans2[j] << " ";
      cout << endl;
    } else {
      cout << "NO\n";
    }
  }
}
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  for (long long i1 = 0; i1 < t; i1++) {
    solve();
  }
  return 0;
}