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>
int N, M, A, B;
char table[60][60];
int rows[60];
int cols[60];
int solve() {
  scanf("%d%d%d%d", &N, &M, &A, &B);
  if (A * N != B * M) {
    puts("NO");
    return 0;
  }
  memset(table, 0, sizeof(table));
  memset(rows, 0, sizeof(rows));
  memset(cols, 0, sizeof(cols));
  std::vector<std::pair<int, int>> sorted_cols;
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < M; ++j) {
      table[i][j] = '0';
    }
    sorted_cols.clear();
    for (int j = 0; j < M; ++j) {
      sorted_cols.push_back(std::make_pair(cols[j], j));
    }
    std::sort(sorted_cols.begin(), sorted_cols.end());
    for (int j = 0; j < A; ++j) {
      int col_index = sorted_cols[j].second;
      int col_value = sorted_cols[j].first;
      if (col_value >= B) {
        puts("NO");
        return 0;
      }
      table[i][col_index] = '1';
      ++rows[i];
      ++cols[col_index];
    }
  }
  puts("YES");
  for (int i = 0; i != N; ++i) {
    puts(table[i]);
  }
  return 0;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    solve();
  }
}
