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;
class BSameParitySummands {
 public:
  void solve(std::istream &in, std::ostream &out) {
    int t;
    in >> t;
    vector<int> v;
    for (int i = 0; i < t; i++) {
      int n, k;
      in >> n >> k;
      int rest = k - 1;
      if (n - rest * 2 > 0 && (n - rest * 2) % 2 == 0) {
        out << "YES\n";
        for (int j = 0; j < k - 1; j++) {
          out << "2 ";
        }
        out << n - (k - 1) * 2 << "\n";
        continue;
      }
      if (n - rest > 0 && (n - rest) % 2 != 0) {
        out << "YES\n";
        for (int j = 0; j < k - 1; j++) {
          out << "1 ";
        }
        out << n - (k - 1) << "\n";
        continue;
      }
      out << "NO\n";
    }
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  BSameParitySummands solver;
  std::istream &in(std::cin);
  std::ostream &out(std::cout);
  solver.solve(in, out);
  return 0;
}