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<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iostream>
#define N 510
using namespace std;
int a[N][N],n,m,h,w;
bool b[N][N];
int main()
{
	scanf("%d%d%d%d",&n,&m,&h,&w);
	for(int i=1;i<=n-h+1;i++)
		for(int j=1;j<=m-w+1;j++)
		{
			if(b[i][j]) continue;
			for(int x=i;x<=i+h-1;x++)
				for(int y=j;y<=j+w-1;y++)
					b[x][y]=1;
			for(int x=i;x<=min(i+2*h-2,n);x++)
				for(int y=j;y<=min(j+2*w-2,m);y++)
					a[x][y]=10;
			a[i+h-1][j+w-1]=-(h*w-1)*10-1;
		}
	int sum=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			sum+=a[i][j];
	if(sum>0)
	{
		printf("Yes\n");
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=m;j++) printf("%d ",a[i][j]);
			printf("\n");
		}
	}
	else printf("No\n");
	return 0;
}