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<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

const int N=505;
const int d=1000;

int n,m,h,w,x,y,v;
int a[N][N];

int main(){
	int i,j;
	scanf("%d%d%d%d",&n,&m,&h,&w);
	x=n/h; y=m/w; v=-h*w;
	if (n%h==0&&m%w==0){
		printf("No\n");
		return 0;
	}
	printf("Yes\n");
	for (i=1;i<=x;i++)
		for (j=1;j<=y;j++) a[i*h][j*w]=v*d+d-1;
	for (i=1;i<=n;i++){
		for (j=1;j<=m;j++)
			if (a[i][j]) printf("%d ",a[i][j]);
		else printf("%d ",d);
		printf("\n");
	}
	return 0;
}