Problem: 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.

Examples

Input

3 3 2 2


Output

Yes
1 1 1
1 -4 1
1 1 1


Input

2 4 1 2


Output

No


Input

3 4 2 3


Output

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

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int H,W,h,w;
int mat[550][550];
long long sum;
int main()
{
	scanf("%d%d%d%d",&H,&W,&h,&w);
	for(int i=h;i<=H;i+=h)
		for(int j=w;j<=W;j+=w)
			mat[i][j]=-4000*h*w+3999;
	for(int i=1;i<=H;++i)
		for(int j=1;j<=W;++j)
			if(!mat[i][j])sum+=4000;
			else sum+=mat[i][j];
	if(sum<=0)
	{
		puts("No");
		return 0;
	}
	puts("Yes");
	for(int i=1;i<=H;++i)
		for(int j=1;j<=W;++j)
		{
			if(!mat[i][j])printf("4000 ");
			else
				printf("%d ",mat[i][j]);
			if(j==W)puts("");
		}
	return 0;
}