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;

using i64 = long long;

const int N = 505;

i64 mx[N][N];

int main() {
#ifdef HOME
	freopen("agc16c.in", "r", stdin);
	freopen("agc16c.out", "w", stdout);
#endif
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	i64 huge;
	int n, m, h, w;

	cin >> n >> m >> h >> w;
	if (n % h == 0 && m % w == 0)
		cout << "No\n";
	else {
		cout << "Yes\n";

		huge = (1e9) - i64(1e9) % (h * w - 1);
		for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			mx[i][j] = ((i - 1) % h == 0 && (j - 1) % w == 0) ? -huge : (huge / (h * w - 1) - 1);

		for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			cout << mx[i][j] << " \n"[j == m]; }

	return 0; }