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: Accepted)

#include <stdio.h>

long long mod=1000000007;
long long modp(long long d,long long p){
    if (p==0){
        return 1 ;
    } else if (p==1){
        return d%mod ;
    } else if (p%2==0){
        return ((modp(d,p/2)%mod)*(modp(d,p/2))%mod)%mod;
    } else {
        return ((modp(d,p/2)%mod)*(modp(d,p/2))%mod)*d%mod;
    }
}

int main(void){
    int h,w,a,b;
    scanf("%d%d%d%d",&h,&w,&a,&b);
    
    long long p[h+w],q[h+w],ans=0,tmp;
    p[0]=1;
    for (int i=1; i<h+w; i++){
        p[i]=(i*p[i-1])%mod;
    }
    q[h+w-1]=modp(p[h+w-1],mod-2);
    for (int i=h+w-2; i>=0; i--){
        q[i]=q[i+1]*(i+1)%mod;
    }
    
    for (int i=b+1; i<=w; i++){
        tmp=p[h-a-1+i-1]%mod;
        tmp=(tmp*q[i-1])%mod;
        tmp=(tmp*q[h-a-1])%mod;
        tmp=(tmp*p[a-1+w-i])%mod;
        tmp=(tmp*q[w-i])%mod;
        tmp=(tmp*q[a-1])%mod;
        ans=(ans+tmp)%mod;
    }
        printf("%lld\n",ans%mod);
	return 0 ;
}