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 Ruby (2.3.3) (judge verdict: Runtime Error)

N = gets.to_i
F_CACHE = [1]
M_CACHE = [1]
MOD = 10 ** 9 + 7

class Integer
  def mod_pow(n, mod)
    x = self
    res = 1

    while n > 0
      res = res * x % mod if n[0] == 1
      x = x * x % mod
      n >>= 1
    end

    res
  end

  def mod_inverse(mod)
    mod_pow(mod - 2, mod)
  end
end

def extgcd(a, b)
  if b.zero?
    [1, 0, a]
  else
    x, y, d = extgcd(b, a % b)

    [y, x - (a / b) * y, d]
  end
end

def mod_inverse(a, m)
  x, y, _ = extgcd(a, m)
  (m + x % m) % m
end

(1..(N + 1)).inject(1) do |res, n|
  v = (res * n) % MOD

  F_CACHE << v
  M_CACHE << v.mod_inverse(MOD)

  v
end

class Integer
  def mod_combination(k, mod)
    return 1 if k.zero?
    return 0 if self < k

    # (self - k + 1..self).inject(:*) / k.factorial
    self.factorial * M_CACHE[k] * M_CACHE[self - k]
    # self.factorial * mod_inverse(k.factorial, MOD) * mod_inverse((self - k).factorial, MOD)
  end

  def permutation(k)
    return 1 if k.zero?

    (self - k + 1..self).inject(:*)
  end

  def factorial
    return F_CACHE[self] if F_CACHE[self]
    return 1 if self.zero?

    (1..self).inject(:*)
  end
end

A = gets.split.map(&:to_i)
counter = Hash.new
l = r = -1

A.each_with_index do |a, i|
  if counter[a]
    l = counter[a]
    r = i
    break
  else
    counter[a] = i
  end
end

1.upto(N + 1) do |k|
  puts ((N + 1).mod_combination(k, MOD) - (l + N - r).mod_combination(k - 1, MOD)) % MOD
end