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;
vector<long long> sieve;
void Sieve(int N) {
  const long long maxn = N;
  sieve.resize(maxn);
  for (long long(i) = 0; (i) < (maxn); ++(i)) sieve[i] = i;
  sieve[1] = -1;
  sieve[0] = -1;
  for (long long(i) = 2; (i) <= (maxn); (i) += (1))
    if (i == sieve[i])
      for (long long j = 2 * i; j < maxn; j += i)
        if (sieve[j] == j) sieve[j] = i;
}
long long extended_GCD(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = extended_GCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
long long power(long long a, long long b, long long m = 1000000007) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long modinv(long long a, long long mod = 1000000007) {
  long long x, y;
  extended_GCD(a, mod, x, y);
  if (x < 0) x += mod;
  return x;
}
void solve() {
  long long n;
  long long m, a, b;
  cin >> n >> m >> a >> b;
  long long arr[n][m];
  memset(arr, 0, sizeof(arr));
  long long j = 0;
  long long k = 0;
  if (a * b > n * m) {
    cout << "NO\n";
    return;
  }
  long long k1 = 0;
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < b; ++j) {
      arr[k1++][i] = 1;
      if (k1 == n) k1 = 0;
    }
  }
  for (int i = 0; i < n; ++i) {
    long long c = 0;
    for (int j = 0; j < m; ++j) {
      c += arr[i][j];
    }
    if (c != a) {
      cout << "NO\n";
      return;
    }
  }
  for (int i = 0; i < m; ++i) {
    long long c = 0;
    for (int j = 0; j < n; ++j) {
      c += arr[j][i];
    }
    if (c != b) {
      cout << "NO\n";
      return;
    }
  }
  cout << "YES\n";
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cout << arr[i][j];
    }
    cout << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
