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 double pi = 2 * acos(0.0);
template <typename T>
T Sqr(T x) {
  T n = x * x;
  return n;
}
template <typename T>
T Pow(T B, T P) {
  if (P == 0) return 1;
  if (P & 1)
    return B * Pow(B, P - 1);
  else
    return Sqr(Pow(B, P / 2));
}
template <typename T>
T Abs(T a) {
  if (a < 0)
    return -a;
  else
    return a;
}
template <typename T>
void Print(T ar[], int a, int b) {
  for (int i = a; i + 1 <= b; i++) cout << ar[i] << " ";
  cout << ar[b] << "\n";
}
template <typename T>
void Print(T ar[], int n) {
  for (int i = 0; i + 1 < n; i++) cout << ar[i] << " ";
  cout << ar[n - 1] << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int test;
  cin >> test;
  for (int tc = 1; tc <= test; tc++) {
    long long n, k;
    cin >> n >> k;
    vector<long long> eve, odd;
    long long sume = 0, sumo = 0;
    for (int i = 1; i < k; i++) {
      sumo += 1;
      sume += 2;
      eve.push_back(2);
      odd.push_back(1);
    }
    long long x = n - sume;
    long long y = n - sumo;
    if (x > 0 && x % 2 == 0 && x + sume == n) {
      cout << "YES\n";
      cout << x << " ";
      for (int i : eve) cout << i << " ";
      cout << "\n";
    } else if (y > 0 && y % 2 == 1 && y + sumo == n) {
      cout << "YES\n";
      cout << y << " ";
      for (int i : odd) cout << i << " ";
      cout << "\n";
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}