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;
int powerm(int x, unsigned int y, int p) {
  int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int power(int x, unsigned int y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  else
    return x * power(x, y / 2) * power(x, y / 2);
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int gcdExtended(int a, int b, int *x, int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int modInverse(int a, int m) {
  int x, y;
  int g = gcdExtended(a, m, &x, &y);
  if (g != 1)
    cout << "Inverse doesn't exist";
  else {
    int res = (x % m + m) % m;
    return res;
  }
}
int vis[1001][1001];
int dx[] = {0, -1, 1, 0};
int dy[] = {1, 0, 0, -1};
int n, m;
int relorder[200001];
bool cmp(int x, int y) { return relorder[x] < relorder[y]; }
void solve() {
  int a, b;
  cin >> n >> m >> a >> b;
  if (n * a != m * b) {
    cout << "NO\n";
    return;
  }
  vector<vector<int> > v(n, vector<int>(m, 0));
  int shift;
  for (int i = 1; i < m; i++)
    if (i * n % m == 0) {
      shift = i;
      break;
    }
  for (int i = 0, d = 0; i < n; i++, d += shift) {
    for (int j = 0; j < a; j++) v[i][(j + d) % m] = 1;
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) cout << v[i][j];
    cout << "\n";
  }
}
int main() {
  int q = 1;
  cin >> q;
  int Q = 1;
  while (q--) {
    solve();
  }
}
