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 PyPy3 (2.4.0) (judge verdict: Accepted)

# -*- coding: utf-8 -*-

H,W,A,B = map(int, input().split())
mod = (10 ** 9 + 7)

# 予め組み合わせ計算に必要な階乗と逆元のテーブルを作っておく
factorial = [1] * (H+W)
factorial[0] = 1
factorial[1] = 1
for i in range(2,H+W):
    factorial[i] = factorial[i-1] * i % mod

inverse = [1] * (H+W)
# フェルマーの小定理から(x! ** mod-2 % mod == x! ** -1 % mod)
# powに第三引数入れると冪乗のmod付計算を高速にやってくれる
inverse[H+W-1] = pow(factorial[H+W-1], mod-2, mod)
for i in range(H+W-2, 0, -1):
    # 最後から戻っていくこのループならH+W回powするより処理が速い
    inverse[i] = inverse[i+1] * (i+1) % mod

# 組み合わせの数だけ返してくれる関数(自作)
def nCr(n, r):
    # 10C7 = 10C3
    r = min(r, n-r)
    # 分子の計算
    numerator = factorial[n]
    # 分母の計算
    denominator = inverse[r] * inverse[n-r] % mod
    return numerator * denominator % mod

h = H - A
w = B + 1
ans = 0
# マスを右上に1つずつずらして、必ず通る場所でパターンを足し合わせていく
while h > 0 and w <= W:
    ans = (ans + nCr(h+w-2, h-1) * nCr(H-h+W-w, H-h)) % mod
    h -= 1
    w += 1

print(ans)