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;
void err(istream_iterator<string> it) { cout << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << " ";
  err(++it, args...);
}
template <typename T>
inline void read(T& x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f = (ch == '-') ? -1 : 1, ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
  x *= f;
}
template <typename T, typename... Args>
inline void read(T& x, Args&... args) {
  read(x), read(args...);
}
constexpr int MAXN = 55, MOD = 1e9 + 7;
int tc, n, m, a, b;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  cin >> tc;
  while (tc--) {
    cin >> n >> m >> a >> b;
    vector<int> cntr(n, a), cntc(m, b);
    vector<vector<int>> ans(n);
    for (int i = 0; i < n; i++) ans[i].resize(m);
    if (a * n != b * m) {
      cout << "NO\n";
      continue;
    }
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < a; j++) ans[i][(j + cnt) % m] = 1;
      cnt += a;
    }
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) cout << ans[i][j];
      cout << "\n";
    }
  }
  return 0;
}
