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;
bool sortt(pair<int, double>& a, pair<int, double>& b) {
  if (a.second > b.second) return 1;
  if (a.second == b.second) {
    return a.first < b.first ? 1 : 0;
  }
  return 0;
}
bool flag = 0;
int BSfindfirst(vector<int> v, int start, int end, vector<bool> marked) {
  int mid = (start + end) / 2;
  if (start >= end) {
    if (marked[v[start]] == 0) flag = 1;
    return v[start];
  }
  if (marked[v[mid]] == 0) {
    flag = 1;
    return BSfindfirst(v, start, mid - 1, marked);
  } else
    return BSfindfirst(v, mid + 1, end, marked);
}
std::vector<unsigned int> generate_primes(unsigned int n) {
  vector<bool> isPrime(n + 1, true);
  std::vector<unsigned int> v;
  isPrime[0] = isPrime[1] = 0;
  for (long long i = 2; i * i <= n; ++i) {
    if (isPrime[i]) {
      for (int j = i * 2; j <= n; j += i) isPrime[j] = 0;
    }
  }
  for (int i = 0; i < (unsigned int)isPrime.size(); ++i) {
    if (isPrime[i]) v.push_back(i);
  }
  return v;
}
int main() {
  int t, n, k;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    if (n % 2 == 0) {
      if (k % 2 == 0 && k <= n) {
        cout << "YES" << endl;
        for (int i = 0; i < (int)(k - 1); ++i)
          cout << "1"
               << " ";
        cout << n - k + 1 << endl;
      } else if (k % 2 == 1 && 2 * k <= n) {
        cout << "YES" << endl;
        for (int i = 0; i < (int)(k - 1); ++i) cout << 2 << " ";
        cout << n - 2 * (k - 1) << endl;
      } else
        cout << "NO" << endl;
    } else {
      if (k % 2 == 1 && k <= n) {
        cout << "YES" << endl;
        for (int i = 0; i < (int)(k - 1); ++i)
          cout << "1"
               << " ";
        cout << n - k + 1 << endl;
      } else
        cout << "NO" << endl;
    }
  }
  return 0;
}