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 int infi = 1e15 + 7;
void printVector(vector<long long int> v) {
  for (long long int i = 0; i < (long long int)v.size(); i++)
    cout << v[i] << " ";
  cout << endl;
}
void printArray(long long int arr[], long long int n) {
  for (long long int i = 0; i < n; i++) cout << arr[i] << ' ';
  cout << endl;
}
set<pair<long long int, long long int>> primeset;
vector<long long int> prime_number;
void SieveOfEratosthenes(long long int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long int p = 2; p <= n; p++)
    if (prime[p]) primeset.insert(make_pair(p, 1));
}
long long modExp(long long base, long long power) {
  if (power == 0) {
    return 1;
  } else {
    long long cur = modExp(base, power / 2);
    cur = cur * cur;
    cur = cur % 1000000007;
    if (power % 2 == 1) cur = cur * base;
    cur = cur % 1000000007;
    return cur;
  }
}
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t = 1;
  cin >> t;
  long long int tt = 0;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    long long int nv = n % 2, kv = k % 2;
    if (nv ^ kv == 0) {
      long long int rem = n - k;
      if (rem >= 0) {
        cout << "YES\n";
        k--;
        while (k--) cout << 1 << " ";
        cout << 1 + rem << '\n';
      } else
        cout << "NO\n";
    } else if (nv == 1 && kv == 0)
      cout << "NO\n";
    else {
      n -= 2 * k;
      if (n < 0) {
        cout << "NO\n";
        continue;
      }
      cout << "YES\n";
      k--;
      while (k--) cout << 2 << ' ';
      cout << 2 + n << '\n';
    }
  }
  return 0;
}