Problem:
Score : 400 points

Problem Statement

We have a large square grid with H rows and W columns.
Iroha is now standing in the top-left cell.
She will repeat going right or down to the adjacent cell, until she reaches the bottom-right cell.

However, she cannot enter the cells in the intersection of the bottom A rows and the leftmost B columns. (That is, there are A×B forbidden cells.) There is no restriction on entering the other cells.

Find the number of ways she can travel to the bottom-right cell.

Since this number can be extremely large, print the number modulo 10^9+7.

Constraints

1 ≦ H, W ≦ 100,000

1 ≦ A < H

1 ≦ B < W

Input

The input is given from Standard Input in the following format:

H W A B

Output

Print the number of ways she can travel to the bottom-right cell, modulo 10^9+7.

Sample Input 1

2 3 1 1

Sample Output 1

2

We have a 2×3 grid, but entering the bottom-left cell is forbidden. The number of ways to travel is two: "Right, Right, Down" and "Right, Down, Right".

Sample Input 2

10 7 3 4

Sample Output 2

3570

There are 12 forbidden cells.

Sample Input 3

100000 100000 99999 99999

Sample Output 3

1

Sample Input 4

100000 100000 44444 55555

Sample Output 4

738162020

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MOD (long long)1000000007

int main() {
	int A, B, H, i, j, W;
	scanf("%d%d%d%d", &H, &W, &A, &B);
	int dp1[H-A][B], dp2[H][W-B];

	dp1[0][0] = 1;
	for (j=0;j<B;j++) dp1[0][j] = 1;
	for (i=1;i<H-A;i++) {
		dp1[i][0]=0;
		for (j=1;j<W;j++) {
			dp1[i][j] = dp1[i-1][j] + dp1[i][j-1];
			dp1[i][j] %= MOD;
		}
//		memmove(dp, dp[1], sizeof(int)*W);
//		memset(dp[1], 0, sizeof(int)*W);
	}
	for (j=0;j<W-B;j++) dp2[0][j] = 1;
	for (i=1;i<H-A;i++) {
		dp2[i][0] = dp1[i][B-1]+dp2[i-1][0];
		for (j=1;j<W-B;j++) {
			dp2[i][j] = dp2[i-1][j] + dp2[i][j-1];
		}
	}
	for (i=H-A;i<H;i++) {
		dp2[i][0] = dp2[i-1][0];
		for (j=1;j<W-B;j++) {
			dp2[i][j] = dp2[i-1][j] + dp2[i][j-1];
		}
	}
	
	printf("%d\n", dp2[H-1][W-B-1]);

	return 0;
}