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;
template <class T, class U>
using Pa = pair<T, U>;
template <class T>
using vec = vector<T>;
template <class T>
using vvec = vector<vec<T>>;
void solve() {
  int N, M, A, B;
  cin >> N >> M >> A >> B;
  vvec<int> ans(N, vec<int>(M));
  int s = 0;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < A; j++) {
      ans[i][(s + j) % M] = 1;
    }
    s = (s + A) % M;
  }
  vec<int> row(N), col(M);
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++)
      if (ans[i][j]) row[i]++, col[j]++;
  bool ok = all_of(row.begin(), row.end(), [&](int x) { return x == A; }) &&
            all_of(col.begin(), col.end(), [&](int x) { return x == B; });
  if (!ok) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    for (int i = 0; i < N; i++)
      for (int j = 0; j < M; j++) cout << ans[i][j] << (j != M - 1 ? "" : "\n");
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}
