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;
inline int read() {
  int a = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) a = (a << 3) + (a << 1) + ch - '0';
  return a * f;
}
void write(int x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar((x % 10) ^ 48);
}
inline void writen(int x) {
  write(x);
  puts("");
}
const int mod = 1e9 + 7;
const int N = 50 + 5;
const int inf = 0x3f3f3f3f;
const int INF = 2147483647;
const double PI = acos(-1);
int n, m[N][N];
int w, h, a, b, t;
signed main() {
  t = read();
  while (t--) {
    w = read();
    h = read();
    a = read();
    b = read();
    if ((w * a) ^ (h * b)) {
      puts("NO");
      continue;
    }
    puts("YES");
    for (int i = 1, lst = 0; i <= w; i++, lst = (lst + a) % h)
      for (int j = lst + 1, k = 1; k <= a; k++, j = j % h + 1) m[i][j] = 1;
    for (int i = 1; i <= w; i++, puts(""))
      for (int j = 1; j <= h; j++) write(m[i][j]);
    memset(m, 0, sizeof(m));
  }
  return 0;
}
