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;
const long long INF = 1e9 + 123, MAXN = 2e5 + 47, MEGAINF = 1e18;
template <class T>
inline istream& operator>>(istream& in, vector<T>& a) {
  for (auto& i : a) in >> i;
  return in;
}
template <class T>
inline ostream& operator<<(ostream& out, vector<T>& a) {
  for (auto& i : a) out << i << " ";
  return out;
}
template <class T, class U>
inline istream& operator>>(istream& in, vector<pair<T, U>>& a) {
  for (auto& i : a) in >> i.first >> i.second;
  return in;
}
template <class T, class U>
inline ostream& operator<<(ostream& out, vector<pair<T, U>>& a) {
  for (auto& i : a) out << i.first << " " << i.second << endl;
  return out;
}
signed main() {
  setlocale(LC_ALL, "rus");
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    if ((n - k + 1) % 2 == 1 && n - k + 1 > 0) {
      cout << "YES\n";
      for (long long i = 0; i < k - 1; ++i) cout << "1 ";
      cout << n - k + 1 << "\n";
    } else if ((n - (k - 1) * 2) % 2 == 0 && n - (k - 1) * 2 > 0) {
      cout << "YES\n";
      for (long long i = 0; i < k - 1; ++i) cout << "2 ";
      cout << n - (k - 1) * 2 << "\n";
    } else
      cout << "NO\n";
  }
}