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>
const long long int mod = 1e9 + 7;
using namespace std;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int imod(long long int a, long long int m) {
  return power(a, m - 2, m);
}
void myfun() {
  long long int n, i, j, m, a, b;
  cin >> n >> m >> a >> b;
  long long int mat[n][m];
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) mat[i][j] = 0;
  vector<pair<long long int, long long int>> arr;
  for (i = 0; i < m; i++) {
    arr.push_back({b, i});
  }
  for (i = 0; i < n; i++) {
    for (j = m - 1; j >= m - a; j--) {
      if (arr[j].first > 0) {
        arr[j].first--;
        mat[i][arr[j].second] = 1;
      } else {
        cout << "NO\n";
        return;
      }
    }
    sort(arr.begin(), arr.end());
  }
  for (i = 0; i < n; i++) {
    long long int nc = 0;
    for (j = 0; j < m; j++)
      if (mat[i][j]) nc++;
    if (nc != a) {
      cout << "NO\n";
      return;
    }
  }
  for (j = 0; j < m; j++) {
    long long int nc = 0;
    for (i = 0; i < n; i++)
      if (mat[i][j]) nc++;
    if (nc != b) {
      cout << "NO\n";
      return;
    }
  }
  cout << "YES\n";
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) cout << mat[i][j];
    cout << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t = 1;
  cin >> t;
  while (t--) myfun();
  return 0;
}
