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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
typedef unsigned long ul;
typedef long long ll;
//typedef pair<int, int> mpair;
using namespace std;

int main() {
	cin.tie(0);
   	ios::sync_with_stdio(false);
   	int h, w, a, b;cin >> h >> w >> a >> b;
    int s[100000][100000];
    //vector<vector<int>> s(h, vector<int>(w, 0));
    rep(i, h){
        rep(j, w){
            s[i][j] = 0;
            if(i == 0 || j == 0){s[i][j] = 1;}
        }
    }
    for(int i = 1;i < h;i++){
        if(i >= h-a){
            for(int j =b;j < w;j++){
                if(j == b){
                    s[i][j] = s[i-1][j];
                }else{
                    s[i][j] = (s[i-1][j]+s[i][j-1])% 100000007;
                }
            }
        }else{
            for(int j = 1;j < w;j++){
                s[i][j] = (s[i-1][j]+s[i][j-1])% 100000007;
            }
        }
    }
    cout << s[h-1][w-1] << endl;

	
	return 0;
}