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: Accepted)

#include <algorithm>
#include <stdio.h>
#define MOD 1000000007
#define ll long long
#define maxf 200000

int f[maxf+1], invf[maxf+1];

inline int lgput (int a, int n) {
    int rez = 1;
    while (n) {
        if (n % 2 == 1)
            rez = (1LL * rez * a) % MOD;
        a = (1LL * a * a) % MOD, n /= 2;
    }
    return rez;
}

inline int combinari (int a, int b) {
    if(b < a) return 0;
    else return 1LL * f[b] % MOD * invf[a] % MOD * invf[b-a] % MOD;
}
int main() {
    int n, m, a, b, nr = 0;
    scanf("%d%d%d%d", &n, &m, &a, &b);

    f[0] = 1;
    invf[0] = 1;
    for (int i = 1; i <= maxf; i++) {
        f[i] = 1LL * i * f[i-1] % MOD;
        invf[i] = 1LL * lgput (f[i], MOD - 2);
    }

    for (int i = 1; i <= n - a; i++) {
        nr = (nr + 1LL * combinari(i - 1, b + i - 2) * combinari(m - b - 1, m + n - b - i - 1) % MOD)% MOD;
    }

    printf("%d", nr);

    return 0;
}