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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val (H, W, h, w) = readLine()!!.split(" ").map {it.toInt()}

    if (H % h == 0 && W % w == 0) {
        println("No")
        return
    }

    val A: Array<Array<Long>>

    if (H % h == 0) {
        val B = func(W, H, w, h)
        A = Array(H) { i -> Array(W)  { j-> B[j][i] } }
    } else {
        A = func(H, W, h, w)
    }

    println("Yes")
    for (r in A) {
        println(r.joinToString(" "))
    }
}

fun func(W: Int, H: Int, w: Int, h: Int): Array<Array<Long>> {
    val V = 1e6.toLong()
    val A = Array(W) { Array(H) { -V - 1L } }

    for (i in 0 until W step w) {
        for (j in 0 until H) {
            A[i][j] = V * (w - 1)
        }
    }

    return A
}