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 testcase() {
  long long n;
  cin >> n;
  int integ;
  cin >> integ;
  vector<long long> v(integ);
  for (int i = 0; i < integ; i++) {
    v[i] = n / integ;
  }
  if (n < integ) {
    std::cout << "NO";
    return;
  }
  n -= integ * v[0];
  if (n % 2 == 0) {
    std::cout << "YES\n";
    v[0] += n;
    for (int i = 0; i < v.size(); i++) {
      std::cout << v[i] << ' ';
    }
  } else if (n % 2 == 1 && integ % 2 == 0) {
    std::cout << "NO";
  } else if (n % 2 == 1 && integ % 2 == 1) {
    if (v[0] - 1 == 0) {
      std::cout << "NO";
      return;
    }
    std::cout << "YES\n";
    v[0] += n + integ;
    for (int i = 0; i < v.size(); i++) {
      std::cout << v[i] - 1 << ' ';
    }
  }
}
int main() {
  int s;
  std::cin >> s;
  for (int i = 0; i < s; i++) {
    testcase();
    std::cout << "\n";
  }
  return 0;
}