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;
using ll = long long;
using P = pair<int, int>;
using vi = vector<int>;
using vc = vector<char>;
using vb = vector<bool>;
using vs = vector<string>;
using vll = vector<long long>;
using vp = vector<pair<int, int>>;
using vvi = vector<vector<int>>;
using vvc = vector<vector<char>>;
using vvll = vector<vector<long long>>;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n, m, a, b;
    cin >> n >> m >> a >> b;
    if (n * a != m * b) {
      cout << "NO" << endl;
      continue;
    }
    vvi ans(n, vi(m, 0));
    for (int i = 0; i < (int)(n); i++) {
      int l = a * i;
      int r = l + a;
      l %= m;
      r %= m;
      bool flag = false;
      if (l >= r) flag = true;
      for (int j = 0; j < (int)(m); j++) {
        if (!flag) {
          if (l <= j && j < r) ans[i][j] = 1;
        } else {
          if (l <= j || j < r) ans[i][j] = 1;
        }
      }
    }
    bool ok = true;
    for (int i = 0; i < (int)(n); i++) {
      int cnt = 0;
      for (int j = 0; j < (int)(m); j++) cnt += ans[i][j];
      if (cnt != a) ok = false;
    }
    for (int j = 0; j < (int)(m); j++) {
      int cnt = 0;
      for (int i = 0; i < (int)(n); i++) cnt += ans[i][j];
      if (cnt != b) ok = false;
    }
    if (!ok) {
      cout << "NO" << endl;
      continue;
    }
    cout << "YES" << endl;
    for (int i = 0; i < (int)(n); i++) {
      for (int j = 0; j < (int)(m); j++) {
        cout << ans[i][j];
      }
      cout << endl;
    }
  }
}
