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 Common Lisp (SBCL 1.1.14) (judge verdict: Runtime Error)

#include <iostream>
using namespace std;
#define M 1000000007

int modexp(int x, int e, int m) {
    long long ans = 1, p = x % m;
    while (e > 0) {
        if (e % 2 != 0) ans = ans * p % m;
        p = p * p % m;
        e >>= 1;
    }
    return ans;
}

long long fact[200001], facti[200001];

long long C(int n, int k) {
    if (n < 0 || k < 0 || k > n) return 0;
    return fact[n] * facti[k] % M * facti[n-k] % M;
}

int main(int argc, char** argv) {
    int h, w, a, b; cin >> h >> w >> a >> b;
    fact[0] = 1; facti[0] = 1;
    for (int i = 1; i <= h+w; i++) {
        fact[i] = fact[i-1] * i % M;
        facti[i] = modexp(fact[i], M-2, M);
    }
    long long ans = 0;
    for (int i = 1; i <= h-a; i++) ans += C(i+b-2, i-1) * C(h+w-b-i-1, h-i) % M;
    cout << ans % M << endl;
}