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 int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const long long N = 11;
const double pai = 3.14159265358979;
const long long maxn2 = 1e3 + 2;
long long read() {
  long long x = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      w = -1;
    }
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return w * x;
}
long long aa[55][55];
int main() {
  long long t = read();
  for (long long h = 1; h <= t; ++h) {
    long long n = read();
    long long m = read();
    long long a = read();
    long long b = read();
    if (n * a != m * b) {
      cout << "NO" << endl;
      continue;
    }
    long long k = n * a;
    for (long long i = 1; i <= n; ++i)
      for (long long j = 1; j <= m; ++j) aa[i][j] = 0;
    long long tmp = 0;
    long long s = 1;
    long long ss = 1;
    while (tmp < k) {
      for (long long i = 1; i <= n; ++i) {
        if (aa[i][s] == 1) {
          s++;
          while (!aa[i][s]) {
            if (aa[i][s] == 0) break;
            s++;
          }
        }
        aa[i][s] = 1;
        tmp++;
        s++;
        if (s > m) s = 1;
      }
    }
    cout << "YES" << endl;
    for (long long i = 1; i <= n; ++i) {
      for (long long j = 1; j <= m; ++j) cout << aa[i][j];
      cout << endl;
    }
  }
  return 0;
}
