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;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long powmod(long long a, long long b, long long mod) {
  long long sum = 1;
  while (b) {
    if (b & 1) {
      sum = (sum * a) % mod;
      b--;
    }
    b /= 2;
    a = a * a % mod;
  }
  return sum;
}
const double Pi = acos(-1.0);
const double epsilon = Pi / 180.0;
const int maxn = 2e5 + 10;
int n, m, a, b;
int t[60][60];
void solve() {
  int now = 0;
  for (int i = 0; i <= n - 1; i++) {
    int tmp = a;
    int tail = now;
    for (int j = 0; j <= tmp - 1; j++) {
      t[i][tail] = 1;
      tail = (tail + 1) % m;
    }
    now = (now + a) % m;
  }
}
int main() {
  int _;
  scanf("%d", &_);
  while (_--) {
    memset(t, 0, sizeof(t));
    cin >> n >> m >> a >> b;
    if (a * n != b * m) {
      cout << "NO" << '\n';
      continue;
    }
    cout << "YES" << '\n';
    int x = 0, y = 0;
    solve();
    for (int i = 0; i <= n - 1; i++) {
      for (int j = 0; j <= m - 1; j++) {
        cout << t[i][j];
      }
      cout << '\n';
    }
  }
}
