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>
bool isT = true;
void solv(int curt);
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  if (isT) {
    int t;
    std::cin >> t;
    for (int i = 1; i <= t; i++) {
      solv(i);
    }
    return 0;
  }
  solv(0);
  return 0;
}
void solv(int curt) {
  long long n, m;
  std::cin >> n >> m;
  if ((n - m + 1LL) % 2LL == 1LL && (n - m + 1) > 0LL) {
    std::cout << "YES\n";
    std::cout << n - m + 1LL << " ";
    for (long long i = 0LL; i < m - 1LL; i++) {
      std::cout << "1"
                << " \n"[i == m - 2LL];
    }
    return;
  }
  if ((n - 2 * (m - 1)) % 2LL == 0LL && n - 2 * (m - 1) > 0LL) {
    std::cout << "YES\n";
    std::cout << n - 2 * (m - 1) << " ";
    for (long long i = 0LL; i < m - 1LL; i++) {
      std::cout << "2"
                << " \n"[i == m - 2LL];
    }
    return;
  }
  std::cout << "NO\n";
}