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;
long long PowModulo(long long a, long long b) {
  long long res = 1;
  a = a % (long long)(1e9 + 7);
  if (b == 0) return 1;
  if (a == 0) return 0;
  while (b) {
    if (b & 1) res = (res * a) % (long long)(1e9 + 7);
    b = b >> 1;
    a = (a * a) % (long long)(1e9 + 7);
  }
  return res;
}
vector<long long> PrefixSum(const vector<long long>& a) {
  vector<long long> ps(a.size() + 1);
  ps[0] = 0;
  for (long long i = 1; i <= a.size(); i++) {
    ps[i] = a[i - 1];
    ps[i] += ps[i - 1];
  }
  return ps;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tc;
  cin >> tc;
  cin.ignore();
  while (tc--) {
    long long n;
    long long k;
    cin >> n >> k;
    long long rem, div;
    div = n / k;
    rem = n % k;
    if (n < k) {
      cout << "NO" << '\n';
      continue;
    }
    if (k == 1) {
      cout << "YES\n" << n << '\n';
      continue;
    }
    if (rem == 0) {
      cout << "YES" << '\n';
      for (long long i = 0; i < k; i++) {
        cout << div << " ";
      }
      cout << '\n';
    } else {
      if (rem % 2 == 0) {
        cout << "Yes" << '\n';
        for (long long i = 0; i < k - 1; i++) {
          cout << div << " ";
        }
        cout << div + rem << '\n';
      } else {
        if (div > 1 && (k - 1) % 2 == 0) {
          cout << "YES" << '\n';
          long long i = 0;
          for (long long i = 0; i < k - 1; i++) {
            long long ans = div;
            if (i % 2)
              ans = div - 1;
            else
              ans = div + 1;
            cout << ans << " ";
          }
          cout << div + rem << '\n';
        } else
          cout << "No" << '\n';
      }
    }
  }
  return 0;
}