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 Python (3.8.2) (judge verdict: Accepted)

import sys

sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
f_inf = float('inf')
mod = 10 ** 9 + 7

class CmbMod:
    def __init__(self, n, p):
        """
        二項係数nCr(n個の区別できるものからr個のものを選ぶ組み合わせの数)をpで割った余りを求める
        """
        self.n = n
        self.p = p
        self.fact = [1, 1]
        self.factinv = [1, 1]
        self.inv = [0, 1]

    def cmb_mod(self, n, r):
        """
        二項係数nCr(mod p)をO(r)にて計算。nが大きいがrは小さい時に使用。
        """
        numer, denom = 1, 1
        for i in range(r):
            numer = (numer * (n - i)) % self.p
            denom = (denom * (i + 1)) % self.p
        return (numer * pow(denom, self.p - 2, self.p)) % self.p

    def prep(self):
        """
        二項係数nCr(mod p)をO(1)で求める為の前処理をO(N)にて実行。
        """
        for i in range(2, self.n + 1):
            self.fact.append((self.fact[-1] * i) % self.p)
            self.inv.append((-self.inv[self.p % i] * (self.p // i)) % self.p)
            self.factinv.append((self.factinv[-1] * self.inv[-1]) % self.p)

    def cmb_mod_with_prep(self, n, r):
        """
        二項係数nCr(mod p)をO(1)で求める。事前にprepを実行する事。
        """
        if (r < 0) or (n < r):
            return 0
        r = min(r, n - r)
        return self.fact[n] * self.factinv[r] * self.factinv[n - r] % self.p


def resolve():
    H, W, A, B = map(int, input().split())
    H -= 1
    W -= 1
    cmb = CmbMod(H + W, mod)
    cmb.prep()
    total = cmb.cmb_mod_with_prep(H + W, H)
    for w in range(B):
        tmp = (cmb.cmb_mod_with_prep(H - A + w, w) * cmb.cmb_mod_with_prep(A - 1 + W - w, A - 1)) % mod
        total -= tmp
        total %= mod
    print(total)


if __name__ == '__main__':
    resolve()