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 <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 (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T1, class T2>
ostream& operator<<(ostream& s, pair<T1, T2> P) {
  return s << '<' << P.first << ", " << P.second << '>';
}
template <class T>
ostream& operator<<(ostream& s, vector<T> P) {
  for (long long i = 0; i < P.size(); ++i) {
    if (i > 0) {
      s << " ";
    }
    s << P[i];
  }
  return s;
}
template <class T>
ostream& operator<<(ostream& s, vector<vector<T> > P) {
  for (long long i = 0; i < P.size(); ++i) {
    s << endl << P[i];
  }
  return s << endl;
}
template <class T>
ostream& operator<<(ostream& s, set<T> P) {
  for (__typeof__((P).begin()) it = (P).begin(); it != (P).end(); ++it) {
    s << "<" << *it << "> ";
  }
  return s << endl;
}
template <class T1, class T2>
ostream& operator<<(ostream& s, map<T1, T2> P) {
  for (__typeof__((P).begin()) it = (P).begin(); it != (P).end(); ++it) {
    s << "<" << it->first << "->" << it->second << "> ";
  }
  return s << endl;
}
template <class T>
void show(vector<T> v) {
  for (long long i = 0; i < v.size(); i++) {
    cerr << v[i] << " ";
  }
  cerr << "\n";
}
void solve() {
  long long n, m, a, b;
  cin >> n >> m >> a >> b;
  if (n * a != m * b) {
    cout << "NO" << endl;
    return;
  }
  long long grid[51][51];
  long long idx = 0;
  for (long long i = 0; i < n; ++i)
    for (long long j = 0; j < m; ++j) grid[i][j] = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = idx; j < a + idx; j++) {
      grid[i][j % m] = 1;
    }
    idx += a;
    idx %= m;
  }
  cout << "YES" << endl;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      cout << grid[i][j];
    }
    cout << endl;
  }
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
