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>
int mbk(long long a,int n,int m){return n?(mbk((a*a)%m,n/2,m)*(n%2?a:1))%m:1;}
int inv(int a,int m){return mbk((long long)a,m-2,m);}
int aCb(int a,int b,int m){
  long long s=1,i;
  if(b>a-b)b=a-b;
  for(i=0;i<b;i++)s=(((s*(a-i))%m)*inv(i+1,m))%m;  
  return (int)s;
}
int main(){
  int h,w,a,b,i,M=1e9+7;
  long long s=0,ans=0,n=1,m=1;
  scanf("%d %d %d %d",&h,&w,&a,&b);
  for(i=0;i<h+w-b-i-2;i++)m=(m*i)%M;
  m=aCb(h+w-b-2,w-b-1,M);//printf("%d %d %lld\n",h+w-b-2,w-b-1,m);
  for(i=0;i<h-a;i++){//printf("%lld %lld\n",n,m);
    s=(n*m)%M;
    ans=(ans+s)%M;
    n=(n*inv(i+1,M))%M;
    n=(n*(b+i))%M;
    m=(m*inv(h+w-i-b-2,M))%M;
    m=(m*(h-i-1))%M;
    //n=(n*aCb(h-i+w-b-2,w-b-1,M))%M;
    //printf("%d %d %d %d %lld\n",b+i-1,i,h-i+w-b-2,w-b-1,n);
    //s=(s+n)%M;//printf("%lld\n",s);
  }
  printf("%lld\n",ans);
  return 0;
}