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 double pi = 3.14159265358979;
const long long int mod = 1e9 + 7;
long long int takemod(long long int a, long long int m = mod) {
  return ((a % m) + m) % m;
}
long long int power(long long int b, long long int p) {
  if (p == 0)
    return 1;
  else if (p % 2 == 0) {
    long long int q = power(b, p / 2);
    return ((q) * (q));
  } else
    return ((b) * (power(b, p - 1)));
}
long long int powerm(long long int b, long long int p) {
  if (p == 0)
    return 1;
  else if (p % 2 == 0) {
    long long int q = powerm(b, p / 2);
    return ((q % mod) * (q % mod)) % mod;
  } else
    return ((b % mod) * (powerm(b, p - 1) % mod)) % mod;
}
bool isprime(long long int n) {
  if (n == 1) return false;
  if (n == 2 or n == 3) return true;
  long long int lim = sqrt(n);
  long long int i = 2;
  while (i <= lim) {
    if (n % i == 0) return false;
    i++;
  }
  return true;
}
vector<long long int> factors(long long int n) {
  vector<long long int> a;
  a.push_back(1);
  long long int lim = sqrt(n);
  long long int i = 2;
  while (i <= lim) {
    if (n % i == 0) {
      a.push_back(i);
      if (i != n / i) a.push_back(n / i);
    }
    i++;
  }
  if (n != 1) a.push_back(n);
  return a;
}
vector<pair<long long int, long long int> > pf_with_count(long long int b) {
  pair<long long int, long long int> p;
  vector<pair<long long int, long long int> > pf;
  long long int count = 0;
  while (b % 2 == 0) {
    b /= 2;
    count++;
  }
  if (count > 0) {
    p.first = 2;
    p.second = count;
    pf.push_back(p);
  }
  long long int i = 3;
  while (i <= sqrt(b)) {
    count = 0;
    while (b % i == 0) {
      b /= i;
      count++;
    }
    if (count > 0) {
      p.first = i;
      p.second = count;
      pf.push_back(p);
    }
    i += 2;
  }
  if (b > 2) {
    p.first = b;
    p.second = 1;
    pf.push_back(p);
  }
  sort(pf.begin(), pf.end());
  return pf;
}
long long int bs_t_i(string second) {
  long long int l = strtoull(second.c_str(), NULL, 2);
  return l;
}
string i_to_bs(long long int n) {
  string second = bitset<7>(n).to_string();
  return second;
}
long long int s_t_i(string second) {
  stringstream obj(second);
  long long int x = 0;
  obj >> x;
  return x;
}
void sol() {
  long long int n, k;
  cin >> n >> k;
  if (n - (k - 1) > 0 and (n - (k - 1)) % 2 != 0) {
    cout << "YES" << endl;
    long long int val = k - 1;
    while (val--) cout << 1 << " ";
    cout << n - k + 1 << endl;
    return;
  }
  if (n - 2 * (k - 1) > 0 and (n - 2 * (k - 1)) % 2 == 0) {
    long long int val = k - 1;
    cout << "YES" << endl;
    while (val--) cout << 2 << " ";
    cout << n - 2 * (k - 1) << endl;
    return;
  }
  cout << "NO" << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  cin >> t;
  while (t--) sol();
}