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;
int gcd(int a, int b, int &x, int &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  int x1, y1;
  int g = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return g;
}
bool diofant_any(int a, int b, int c, int &x, int &y, int &g) {
  g = gcd(abs(a), abs(b), x, y);
  if (c % g != 0) return false;
  x *= c / g;
  y *= c / g;
  if (a < 0) x *= -1;
  if (b < 0) y *= -1;
  return true;
}
int y;
vector<int> sifr(int n) {
  vector<int> res;
  while (n) {
    if (n % 10) res.push_back(n % 10);
    n /= 10;
  }
  return res;
}
int main() {
  int t, n, k;
  cin >> t;
  for (int tt = 0; tt < t; ++tt) {
    cin >> n >> k;
    if ((n % 2 == 1 && k % 2 == 0) || (n % 2 == k % 2 && n - k + 1 <= 0) ||
        (n % 2 == 0 && k % 2 == 1 && n - 2 * k + 2 <= 0))
      cout << "NO\n";
    else if (n % 2 == k % 2) {
      cout << "YES\n";
      for (int i = 0; i < k - 1; ++i) {
        cout << 1 << ' ';
      }
      cout << n - k + 1 << '\n';
    } else if (n % 2 == 0 && k % 2 == 1) {
      cout << "YES\n";
      for (int i = 0; i < k - 1; ++i) {
        cout << 2 << ' ';
      }
      cout << n - 2 * k + 2 << '\n';
    }
  }
  return 0;
}