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: Accepted)

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <numeric>
#include <functional>

using namespace std;

vector<int> fact, ifact;
const int mod = 1000000007;

vector<int> range_mod_inverse(int n, const int mod) {
  vector<int> ret(n + 1);
  ret[1] = 1;
  for (int i = 2; i <= n; i++)
    ret[i] = (long long)(mod - mod / i) * ret[mod%i] % mod;
  return ret;
}

vector<int> getFactorials(int n, const int mod) {
  vector<int> ret(1, 1);
  for (int i = 1; i <= n; i++) ret.push_back((long long)ret.back()*i%mod);
  return ret;
}

vector<int> getInverseFactorials(int n, const int mod) {
  vector<int> invmod(range_mod_inverse(n, mod));
  vector<int> ret(1, 1);
  for (int i = 1; i <= n; i++) ret.push_back((long long)ret.back()*invmod[i] % mod);
  return ret;
}

long long binomial(long long n, long long r) {
  if (n < 0 || r < 0 || r > n)
    return 0;
  return (long long)fact[n] * ifact[r] % mod * ifact[n-r] % mod;
}

int main()
{
  fact = getFactorials(200005, mod);
  ifact = getInverseFactorials(200005, mod);
  int h, w, a, b;
  scanf("%d%d%d%d", &h, &w, &a, &b);
  long long ans = binomial(h-1+w-1, w-1);
  for (int i = b - 1; i >= 0; i--) {
    int j = a - 1 - (b - 1 - i);
    if (j < 0)
      break;
    int x = i;
    int y = h - j - 1;
    ans -= binomial(x + y, x) * binomial(w-1-x + h-1-y, w-1-x) % mod;
  }
  ans %= mod;
  ans += mod;
  ans %= mod;
  printf("%lld\n", ans);
  return 0;
}