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 <bits/stdc++.h>
using namespace std;
#define FOR(i, n) for(int i = 0; i < (n); i++)
#define MEM(a, x) memset(a, x, sizeof(a))
#define ALL(a) a.begin(), a.end()
#define UNIQUE(a) a.erase(unique(ALL(a)), a.end())
typedef long long ll;

int H, W, h, w;
ll a[505][505];

int main(int argc, char const *argv[]) {
	ios_base::sync_with_stdio(false);
	cin >> H >> W >> h >> w;
	ll sum = 0;
	for (int i = 1; i <= H; i++) {
		for (int j = 1; j <= W; j++) {
			if (i % h == 0 && j % w == 0) a[i][j] = -h*w;
			else a[i][j] = 1;
			sum += (ll)a[i][j];
		}
	}
	if (sum > 0) {
		cout << "Yes" << endl;
		for (int i = 1; i <= H; i++) {
			for (int j = 1; j <= W; j++) {
				cout << a[i][j] << (j == W ? "\n" : " ");
			}
		}
	}
	else cout << "No" << endl;
	return 0;
}