You are given four positive integers n, m, a, b (1 ≤ b ≤ n ≤ 50; 1 ≤ a ≤ m ≤ 50). Find any such rectangular matrix of size n × m that satisfies all of the following conditions:

  * each row of the matrix contains exactly a ones; 
  * each column of the matrix contains exactly b ones; 
  * all other elements are zeros. 



If the desired matrix does not exist, indicate this.

For example, for n=3, m=6, a=2, b=1, there exists a matrix satisfying the conditions above:

$$$ \begin{vmatrix} 0 & 1 & 0 & 0 & 0 & 1 \\\ 1 & 0 & 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 & 1 & 0 \end{vmatrix} $$$

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.

Each test case is described by four positive integers n, m, a, b (1 ≤ b ≤ n ≤ 50; 1 ≤ a ≤ m ≤ 50), where n and m are the sizes of the matrix, and a and b are the number of ones for rows and columns, respectively.

Output

For each test case print:

  * "YES" (without quotes) and the required matrix (if there are several answers, print any) if it exists, or 
  * "NO" (without quotes) if it does not exist. 



To print the matrix n × m, print n rows, each of which consists of m numbers 0 or 1 describing a row of the matrix. Numbers must be printed without spaces.

Example

Input


5
3 6 2 1
2 2 2 1
2 2 2 2
4 4 2 2
2 1 1 2


Output


YES
010001
100100
001010
NO
YES
11
11
YES
1100
1100
0011
0011
YES
1
1
#include <bits/stdc++.h>
using namespace std;
vector<int> pr;
void sieve(int n) {
  bool a[n + 1];
  memset(a, true, sizeof(a));
  for (int i = 2; i * i <= n; i++) {
    if (a[i]) {
      for (int j = i * i; j <= n; j += i) a[j] = false;
    }
    if (a[i]) pr.push_back(i);
  }
}
long long kmp(string s) {
  long long sz = (long long)s.size();
  vector<long long> lps(sz, 0);
  for (long long i = 1; i < sz; i++) {
    long long pred = lps[i - 1];
    while (pred > 0 && s[i] != s[pred]) {
      pred = lps[pred - 1];
    }
    lps[i] = pred + (s[i] == s[pred] ? 1 : 0);
  }
  return lps[sz - 1];
}
bool isp(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void solve() {
  int n, m, r, c;
  cin >> n >> m >> r >> c;
  if (n * r != m * c)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    int a[n][m];
    memset(a, 0, sizeof(a));
    vector<pair<int, int> > v(m);
    for (long long i = 0; i < m; i++) {
      v[i].first = 0;
      v[i].second = i;
    }
    for (long long i = 0; i < n; i++) {
      for (int j = 0; j < r; j++) {
        sort(v.begin(), v.end());
        v[0].first = v[0].first + 1;
        a[i][v[0].second] = 1;
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        cout << a[i][j];
      }
      cout << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
