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;
template <typename Arg1, typename Arg2>
ostream& operator<<(ostream& out, const pair<Arg1, Arg2>& x) {
  return out << "(" << x.first << "," << x.second << ")";
}
template <typename Arg1>
ostream& operator<<(ostream& out, const vector<Arg1>& a) {
  out << "[";
  for (const auto& x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1>
ostream& operator<<(ostream& out, const set<Arg1>& a) {
  out << "[";
  for (const auto& x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1, typename Arg2>
ostream& operator<<(ostream& out, const map<Arg1, Arg2>& a) {
  out << "[";
  for (const auto& x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1>
void __trace__(const string name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << " ] " << endl;
}
template <typename Arg1, typename... Args>
void __trace__(const string names, Arg1&& arg1, Args&&... args) {
  const string name = names.substr(0, names.find(','));
  cerr << name << " : " << arg1 << " | ";
  __trace__(names.substr(1 + (int)name.size()), args...);
}
clock_t time_p = clock();
void time_taken() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
const long long mod = 1e9 + 7;
const long long INF = 1e18;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(12);
  cout << fixed;
  int t;
  cin >> t;
  while (t--) {
    int n, m, a, b;
    cin >> n >> m >> a >> b;
    if (a * n != m * b) {
      cout << "NO" << '\n';
      continue;
    }
    vector<vector<int> > ans(n, vector<int>(m, 0));
    for (int i = 0; i < n; i++) {
      int cnt = 0;
      for (int j = (a * i) % m; cnt < a; j = (j + 1) % m) {
        ans[i][j] = 1;
        cnt++;
      }
    }
    cout << "YES" << '\n';
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        cout << ans[i][j];
      }
      cout << '\n';
    }
  }
  time_taken();
  return 0;
}
