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<iostream>
#include<cstdio>
#include<cstring>

const int N=510;

int G[N][N];

void solve(int n,int m,int dn)
{
	int B0=(n/dn)*m,A0=n*m-B0;
	int a=A0-B0*(dn-1),K=A0/a+1;

	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(i%dn==0)G[i][j]=-K*(dn-1);
			else G[i][j]=K-1;
}

int n,m,dn,dm;

int main()
{
	scanf("%d%d%d%d",&n,&m,&dn,&dm);
	if(n%dn==0 && m%dm==0)printf("No\n");
	else 
	{
		printf("Yes\n");
		if(n%dn==0)
		{
			solve(m,n,dm);
			for(int i=1;i<=n;i++)
				for(int j=1;j<=m;j++)
					std::swap(G[i][j],G[j][i]);
		}
		else solve(n,m,dn);
		
		for(int i=1;i<=n;i++,printf("\n"))
			for(int j=1;j<=m;j++)printf("%d ",G[i][j]);
	}
	return 0;
}