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 Rust (1.15.1) (judge verdict: Memory Limit Exceeded)

use std::io;
use std::io::Read;

fn main() {
    let mut stdin = io::stdin();
    let mut lines = String::new();
    let _ = stdin.read_to_string(&mut lines);
    let vec = lines.lines().collect::<Vec<&str>>();
    let vec2 = vec[0].split(' ').collect::<Vec<&str>>();
    let h :usize = vec2[0].trim().parse().unwrap();
    let w :usize = vec2[1].trim().parse().unwrap();
    let a :usize = vec2[2].trim().parse().unwrap();
    let b :usize = vec2[3].trim().parse().unwrap();
    let mo = 1000000007;

    let mut dp :Vec<usize> = vec![0;h*w];
    for i in 0..w {
        for j in 0..h {
            if i<b && j+a>=h {
                dp[j*w+i] = 0;
            } else if i==0 || j==0 {
                dp[j*w+i] = 1;
            } else {
                dp[j*w+i] = ( dp[(j-1)*w+i] + dp[j*w+i-1] ) % mo;
            }
        }
    }
    println!("{}", dp[h*w-1]);
}