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;
long long mul(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  return (a % p + b % p) % p;
}
void input(long long a[], long long sz) {
  for (long long i = 0; i < sz; i++) cin >> a[i];
}
void print(long long a[], long long sz) {
  for (long long i = 0; i < sz; i++) {
    if (i == sz - 1)
      cout << a[i] << "\n";
    else
      cout << a[i] << " ";
  }
}
long long maxr(long long a[], long long sz) {
  long long ma;
  for (long long i = 0; i < sz; i++) {
    if (i == 0)
      ma = a[i];
    else if (a[i] > ma)
      ma = a[i];
  }
  return ma;
}
long long minr(long long a[], long long sz) {
  long long mi;
  for (long long i = 0; i < sz; i++) {
    if (i == 0)
      mi = a[i];
    else if (a[i] < mi)
      mi = a[i];
  }
  return mi;
}
long long isprm(long long n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  return 1;
}
long long power(long long x, long long y, long long p = 1000000007) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, long long p = 1000000007) {
  return power(n, p - 2, p);
}
long long ncrMod(long long n, long long r, long long p = 1000000007) {
  if (r == 0) return 1;
  long long fac[n + 1];
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
bool comp(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first == b.first) {
    return (a.second > b.second);
  }
  return (a.first < b.first);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long q = n / k;
    long long r = n % k;
    if ((r % 2 == 1 && k % 2 == 0) || (r % 2 == 1 && q == 1) || q == 0)
      cout << "NO\n";
    else {
      cout << "YES\n";
      if (r % 2 == 0) {
        cout << q + r << " ";
        for (long long i = 0; i < k - 1; i++) cout << q << " ";
        cout << "\n";
      } else {
        cout << q + r << " ";
        for (long long i = 0; i < k - 1; i++) {
          if (i % 2 == 0)
            cout << q - 1 << " ";
          else
            cout << q + 1 << " ";
        }
        cout << "\n";
      }
    }
  }
  return 0;
}