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 sc1() {
  int x;
  scanf("%d", &x);
  return x;
}
long long sc2() {
  long long x;
  scanf("%lld", &x);
  return x;
}
const int mod = 1e9 + 7;
inline int add(int a, int b) {
  a += b;
  return a >= mod ? a - mod : a;
}
inline int sub(int a, int b) {
  a -= b;
  return a < 0 ? a + mod : a;
}
inline int mul(int a, int b) { return (long long)a * b % mod; }
void err(istream_iterator<string> it) { cout << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << ", ";
  err(++it, args...);
}
const int inf = (int)2e9 + 5;
const long long Inf = (long long)1e18 + 5;
const int N = 5e6 + 5;
int solve() {
  int n = sc1(), k = sc1();
  if (n % 2 == 0) {
    if (k % 2 == 0) {
      if (n < k)
        puts("NO");
      else {
        puts("YES");
        for (int i = 1; i < k; i++) cout << 1 << " ";
        cout << n - (k - 1) << endl;
      }
    } else {
      if (n < (k * 2)) {
        puts("NO");
        return 0;
      } else {
        puts("YES");
        for (int i = 1; i < k; i++) {
          cout << 2 << " ";
        }
        cout << (n - ((k - 1) * 2)) << endl;
      }
    }
  } else {
    if (k % 2 == 0) {
      puts("NO");
      return 0;
    }
    if (n < k)
      puts("NO");
    else {
      puts("YES");
      for (int i = 1; i < k; i++) {
        cout << 1 << " ";
      }
      cout << (n - (k - 1)) << endl;
    }
  }
  return 0;
}
int main() {
  int test = 1, tc = 0;
  scanf("%d", &test);
  while (test--) {
    solve();
  }
  return 0;
}