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 royale_paradox() {
  long long int n, k;
  cin >> n >> k;
  if (n > k) {
    if (n % 2 == 0 && k % 2 == 0) {
      cout << "YES"
           << "\n";
      for (long long int i = 0; i < k - 1; i++) {
        cout << "1"
             << " ";
        n = n - 1;
      }
      cout << n;
    } else if (n % 2 == 0 && k % 2 == 1) {
      if (n >= 2 * k) {
        cout << "YES"
             << "\n";
        for (long long int i = 0; i < k - 1; i++) {
          cout << "2"
               << " ";
          n = n - 2;
        }
        cout << n;
      } else {
        cout << "NO";
      }
    } else if (n % 2 == 1 && k % 2 == 0) {
      cout << "NO";
    } else if (n % 2 == 1 && k % 2 == 1) {
      cout << "YES"
           << "\n";
      for (long long int i = 0; i < k - 1; i++) {
        cout << "1"
             << " ";
        n = n - 1;
      }
      cout << n;
    }
  } else if (n == k) {
    cout << "YES"
         << "\n";
    for (long long int i = 0; i < k; i++) {
      cout << "1"
           << " ";
    }
  } else {
    cout << "NO";
  }
  cout << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    royale_paradox();
  }
  return 0;
}
bool sortbysec(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
bool isSubSequence(string s, string t, int m, int n) {
  if (m == 0) return true;
  if (n == 0) return false;
  if (s[m - 1] == t[n - 1]) return isSubSequence(s, t, m - 1, n - 1);
  return isSubSequence(s, t, m, n - 1);
}