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;
string abc = "abcdefghijklmnopqrstuvwxyz";
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long fact(long long n) {
  long long res = 1;
  for (long long z = 2; z <= n; z++) res = res * z;
  return res;
}
long long Pow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return res;
}
long long amax(long long a[], long long n) {
  long long max = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] > max) max = a[i];
  }
  return max;
}
long long amin(long long a[], long long n) {
  long long min = 99999999999999999;
  for (long long i = 0; i < n; i++) {
    if (a[i] < min) min = a[i];
  }
  return min;
}
void ip(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cin >> a[i];
}
void op(long long a[], long long n) {
  for (long long i = 0; i < n; i++) {
    if (i == n - 1)
      cout << a[i] << endl;
    else
      cout << a[i] << " ";
  }
}
long long divisor(long long x) {
  long long cnt = 0;
  for (int i = 1; i <= sqrt(x); i++) {
    if (x % i == 0) {
      if (i != x / i)
        cnt += 2;
      else
        cnt += 1;
    }
  }
  return cnt;
}
bool palindrome(long long le, long long ri, string s) {
  if (ri - le <= 0) return true;
  if (s[le] != s[ri]) return false;
  return palindrome(le + 1, ri - 1, s);
}
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;
}
bool psqr(long double x) {
  long double sr = sqrt(x);
  return ((sr - floor(sr)) == 0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    std::vector<long long> v;
    cin >> n >> k;
    if (n == k) {
      cout << "YES\n";
      for (long long i = 0; i < k; i++) cout << 1 << " ";
      cout << endl;
    } else if (n - (k - 1) % 2 != 0 && (k - 1) % 2 != 0 && n % 2 == 0 &&
               k - 1 + (n - (k - 1)) == n && (n - (k - 1)) > 0) {
      cout << "YES\n";
      for (long long i = 0; i < k - 1; i++) cout << 1 << " ";
      cout << n - (k - 1) << endl;
    } else if (n < 2 * k && n % 2 == 0)
      cout << "NO\n";
    else if (n < k)
      cout << "NO\n";
    else if (n % 2 != 0 && k % 2 == 0)
      cout << "NO\n";
    else {
      cout << "YES\n";
      if (n % 2 == 0) {
        for (long long i = 0; i < k - 1; i++) v.push_back(2);
        v.push_back(n - ((k - 1) * 2));
      } else {
        for (long long i = 0; i < k - 1; i++) v.push_back(1);
        v.push_back(n - (k - 1));
      }
      for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
      cout << endl;
    }
  }
  return (0);
}