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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(){
	int H, W, h, w, i, j;
	scanf("%d%d%d%d", &H, &W, &h, &w);
	if((H * W) - ((H / h) * (W / w) * (h * w + 1)) <= 0){
		if(h == 1 && w != 1 && W % w == 1){
			printf("Yes\n");
			for(i = 1; i <= H; i++){
				for(j = 1; j <= W; j++){
					if(j % w == 0){
						printf("%d ", - (h * w + 1));
					}
					else if(j % w == 1){
						printf("2 ");
					}
					else{
						printf("1 ");
					}
				}
				printf("\n");
			}
		}
		else if(h != 1 && w == 1 && H % h == 1){
			printf("Yes\n");
			for(i = 1; i <= H; i++){
				for(j = 1; j <= W; j++){
					if(i % h == 0){
						printf("%d ", - (h * w + 1));
					}
					else if(i % h == 1){
						printf("2 ");
					}
					else{
						printf("1 ");
					}
				}
				printf("\n");
			}
		}
		else{
			printf("No\n");
		}
	}
	else{
		printf("Yes\n");
		for(i = 1; i <= H; i++){
			for(j = 1; j <= W; j++){
				if(i % h == 0 && j % w == 0){
					printf("%d ", - h * w);
				}
				else{
					printf("1 ");
				}
			}
			printf("\n");
		}
	}
	return 0;
}