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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long double PI = acos(-1);
const int N = 1e5 + 10;
const int mod = 1e9 + 7;
const int mod1 = 998244353;
const long long INF = 2e18;
template <class T>
inline void amin(T& x, const T& y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T& x, const T& y) {
  if (x < y) x = y;
}
int read() {
  long long s = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    s = s * 10 + c - '0';
    c = getchar();
  }
  return s * f;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    if ((n - (k - 1)) % 2 != 0 && (n - (k - 1)) > 0) {
      cout << "YES"
           << "\n";
      for (int i = 0; i < k - 1; i++) {
        cout << "1 ";
      }
      cout << (n - (k - 1));
      cout << "\n";
      continue;
    }
    if ((n - 2 * (k - 1)) % 2 == 0 && (n - 2 * (k - 1)) > 0) {
      cout << "YES"
           << "\n";
      for (int i = 0; i < k - 1; i++) {
        cout << "2 ";
      }
      cout << (n - 2 * (k - 1));
      cout << "\n";
      continue;
    }
    cout << "NO";
    cout << "\n";
  }
  return 0;
}