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;
using ll = long long;
using ld = long double;
int main() {
  ll t, n, m, a, b;
  cin >> t;
  for (int i = 0; i < t; i += 1) {
    cin >> n >> m >> a >> b;
    if (a > m || b > n) {
      cout << "NO" << endl;
      continue;
    }
    vector<vector<int> > mat(n, vector<int>(m, 0));
    vector<pair<long long, long long> > colCount(m);
    for (int j = 0; j < m; j += 1) {
      colCount[j] = {0, j};
    }
    bool ok = true;
    for (int j = 0; j < n; j += 1) {
      sort(colCount.begin(), colCount.end());
      for (int k = 0; k < a; k += 1) {
        mat[j][colCount[k].second] = 1;
        colCount[k].first++;
        if (colCount[k].first > b) {
          ok = false;
          break;
        }
      }
    }
    for (auto x : colCount) {
      if (x.first != b) {
        ok = false;
        break;
      }
    }
    if (ok) {
      cout << "YES" << endl;
      for (int j = 0; j < n; j += 1) {
        for (int k = 0; k < m; k += 1) {
          cout << mat[j][k];
        }
        cout << endl;
      }
    } else {
      cout << "NO" << endl;
    }
  }
}
