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>
using namespace std;
int b[505][505];
int main(){
	int a[10][2];
	int i,j,k;
	int n,m;
	cin>>a[0][0]>>a[0][1];
	cin>>a[1][0]>>a[1][1];
	a[2][0]=a[1][0]*2-1,a[2][1]=a[1][1]*2-1;
	n=((a[0][0]-1)/a[2][0]+1)*((a[0][1]-1)/a[2][1]+1);
	n=a[0][0]*a[0][1]-n*(a[1][0]*a[1][1]+1);
	if(n<=0){
		cout<<"No"<<endl;
	}else{
		cout<<"Yes"<<endl;
		for(i=0;i<a[0][0];i++){
			for(j=0;j<a[0][1];j++){
				b[i][j]=1;
			}
		}
		for(i=a[1][0]-1;i<a[0][0];i+=a[2][0]){
			for(j=a[1][1]-1;j<a[0][1];j+=a[2][1]){
				b[i][j]=-a[1][0]*a[1][1];
			}
		}
		for(i=0;i<a[0][0];i++){
			for(j=0;j<a[0][1];j++){
				printf("%d%c",b[i][j],j!=a[0][1]-1?' ':'\n');
			}
		}
	}
	return 0;
}