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;
const double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230;
void swaps(char *x, char *y) {
  char temp;
  temp = *x;
  *x = *y;
  *y = temp;
}
void swapi(int *a, int *b) {
  int temp;
  temp = *a;
  *a = *b;
  *b = temp;
}
unsigned long long gcd(unsigned long long a, unsigned long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == 1 || b == 1) return 1;
  if (a == b) return a;
  if (a > b)
    return gcd(b, a % b);
  else
    return gcd(a, b % a);
}
void preprocess() {}
void refresh() {}
void compute() {
  int a, b, m, n, diff, x;
  int matrix[50][50] = {0};
  bool reversed = false, flag = false;
  cin >> n >> m >> a >> b;
  if (n * a != m * b) {
    printf("NO\n");
    return;
  }
  if (n == 1 && m == 1 && a == 1 && b == 1) {
    printf("YES\n1\n");
    return;
  }
  if (n > m) {
    x = n;
    n = m;
    m = x;
    x = a;
    a = b;
    b = x;
    reversed = true;
  }
  printf("YES\n");
  for (diff = 1; diff < m; ++diff) {
    if ((diff * n) % m == 0) {
      break;
    }
  }
  for (int i = 0; i < n; ++i) {
    x = 0;
    for (int j = i * diff; x < a; ++j) {
      j %= m;
      matrix[i][j] = 1;
      x++;
    }
  }
  if (reversed) {
    for (int j = 0; j < m; ++j) {
      for (int i = 0; i < n; ++i) {
        printf("%d", matrix[i][j]);
      }
      printf("\n");
    }
  } else {
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
        printf("%d", matrix[i][j]);
      }
      printf("\n");
    }
  }
}
int main() {
  int t, i, j;
  preprocess();
  scanf("%d", &t);
  while (t--) {
    compute();
  }
  return 0;
}
