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: Wrong Answer)

#include<iostream>

using namespace std;

const int MAX = 510000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}


int main()
{
    int H,W,A,B;
    cin >> H >> W >> A >> B;
    COMinit();
    long long int ans = 0;
    for (int i=1; i<W-B+1; ++i)
    {
        // long long tmp = fac[H-A-1+B+i-1] * (finv[H-A-1] * finv[B+i-1] % MOD) % MOD;
        // tmp *= fac[A-1+W-B-i] * (finv[A-1] * finv[W-B-i] % MOD) % MOD;
        // tmp %= MOD;
        // ans %= MOD;
        // ans += tmp;
        long long tmp = COM(H-A-1+B+i-1,B+i-1)%MOD * COM(A-1+W-B-i,W-B-i)%MOD;
        tmp %= MOD;
        ans %= MOD;
        ans += tmp;

    }
    cout << ans << endl;
    return 0;
}