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>
#pragma GCC diagnostic ignored "-Woverflow"
using namespace std;
const short INF16 = ~(1 << 15);
const int INF32 = 1e9;
const int SZ = (1 << 21);
const int SZ2 = 200;
const long long mod = 998244353;
const long long INF64 = 8e18;
const long long P = 53;
const long double eps = 1e-10;
const long double PI = acosl(-1);
inline long long MOD(long long x) {
  x %= mod, x += mod;
  return x % mod;
}
inline long long MOD(long long x, long long m) {
  x %= m, x += m;
  return x % m;
}
inline long long T(long long x) { return x * (x + 1) >> 1; }
long long fpow(long long a, long long b) {
  if (!b) return 1;
  if (b == 1) return a;
  if (b & 1) return fpow(a, b ^ 1) * a % mod;
  long long d = fpow(a, b >> 1);
  return d * d % mod;
}
void Solve(int num_test) {
  int n, m, a, b;
  cin >> n >> m >> a >> b;
  if (n * a != m * b) {
    cout << "NO" << endl;
    return;
  }
  vector<vector<char>> was(n, vector<char>(m));
  vector<int> cnt(m);
  for (int i = 0; i < a; i++) {
    for (int j = 0; j < n; j++) was[j][i] = 1;
    cnt[i] = n;
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      if (!was[j][i]) continue;
      if (cnt[i] <= b) break;
      for (int k = i + 1; k < m; k++)
        if (!was[j][k]) {
          was[j][k] = 1;
          was[j][i] = 0;
          cnt[i]--;
          cnt[k]++;
          break;
        }
    }
  }
  cout << "YES" << endl;
  for (auto &i : was) {
    for (int j : i) cout << j;
    cout << endl;
  }
}
signed main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(10);
  srand(time(NULL));
  int test = 1;
  cin >> test;
  for (int i = 1; i <= test; i++) Solve(i);
}
