Problem:
Score : 700 points

Problem Statement

You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W).
Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:

The matrix has H rows and W columns.

Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).

The sum of all the elements of the matrix is positive.

The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.

Constraints

1 ≤ h ≤ H ≤ 500

1 ≤ w ≤ W ≤ 500

Input

Input is given from Standard Input in the following format:

H W h w

Output

If there does not exist a matrix that satisfies all of the conditions, print No.

Otherwise, print Yes in the first line, and print a matrix in the subsequent lines in the following format:

a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}

Here, a_{ij} represents the (i,\ j) element of the matrix.

Sample Input 1

3 3 2 2

Sample Output 1

Yes
1 1 1
1 -4 1
1 1 1

The sum of all the elements of this matrix is 4, which is positive.
Also, in this matrix, there are four subrectangles with 2 rows and 2 columns as shown below. For each of them, the sum of all the elements inside is -1, which is negative.

Sample Input 2

2 4 1 2

Sample Output 2

No

Sample Input 3

3 4 2 3

Sample Output 3

Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <cstdio>
using namespace std;

int H, W, h, w;

int main(void) {
  scanf("%d%d%d%d", &H, &W, &h, &w);
	if ((H%h == 0) && (W%w == 0)) {printf("No\n"); return 0;}
	printf("Yes\n");

  int base = 1;
  int b[1001] = {};

  for (int j = W; j > (w - 2); --j) {
    if (b[j] == 0) b[j] = base;
    b[j - w] = b[j] + base;
  }
  int k = 0; int plus = -b[0];
  while(k < W) {
    b[k] = b[k] + plus;
    k += w;
  }

  int c[1001] = {};
  for (int i = 0; i < W; ++i) {
    c[i] = b[i + 1] - b[i];
  }

  for (int i = 0; i < H; ++i) {
    for (int j = 0; j < W; ++j) {
      printf("%d ", c[j]);
    }
    printf("\n");
  }
}