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 fastpow(long long a, long long b,
                  long long m = (long long)(1e9 + 7)) {
  long long res = 1;
  a %= m;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n, m, a, b;
    cin >> n >> m >> a >> b;
    vector<pair<long long, long long>> cols(m);
    vector<vector<long long>> res(n, vector<long long>(m, 0));
    for (long long i = 0; i < m; i++) {
      cols[i].second = i;
    }
    bool poss = true;
    for (long long i = 0; i < n; i++) {
      sort(cols.begin(), cols.end());
      for (long long j = 0; j < a; j++) {
        cols[j].first++;
        if (cols[j].first > b) poss = false;
        res[i][cols[j].second] = 1;
      }
    }
    for (auto c : cols) {
      if (c.first != b) poss = false;
    }
    if (poss) {
      cout << "YES" << endl;
      for (auto r : res) {
        for (auto x : r) cout << x;
        cout << endl;
      }
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}
