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: Wrong Answer)

//In the name of ALLAH the most merciful the most beneficial

#include<cstdio>
#include<cstring>
#include<bits/stdc++.h>
#include<iostream>
#include<math.h>
#include<fstream>
#define sc(x) scanf("%d",&x)
#define ll long long
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define FORd(i,a,b) for(int i=a;i>b;i--)
#define ff first
#define ss second
#define sz(x) x.size()
#define pb push_back

using namespace std;

typedef pair<int,int> pii;
typedef vector<pii> vii;
const int con = 1e5+7;
static const int INF = 0x3f3f3f3f;
static const int MOD = 1e9+7;


int main(){
/*
    clock_t start,end;
    start = clock();*/
    ll H,W,A,B;cin >> H >> W >> A >> B;

    ll dp1[con],dp2[con];
    FOR(i,0,con)
        dp1[i] = 1;
    dp2[0] = 1;



    if(H-A <= 1)
        dp2[0] = 0;
    else
        dp2[0] = 1;

    //H rows W cols
    FOR(i,1,H){


        if(i%2 == 0){
            FOR(j,1,W){
                if(i >= H-A && j < B)
                    dp1[j] = 0;
                else
                    dp1[j] = (dp2[j] + dp1[j-1])%MOD;
            }
        }
        else{
            FOR(j,1,W){
                if(i >= H-A && j< B)
                    dp2[j] = 0;
                else
                    dp2[j] = (dp1[j] + dp2[j-1])%MOD;
            }
        }


    }




    if(H%2 == 0)
        cout << dp2[W-1]%MOD << endl;
    else
        cout << dp1[W-1]%MOD << endl;

/*    end = clock();
    double time = double(end-start)/(double)CLOCKS_PER_SEC;
    cout <<fixed<< time/1000 << "sec";*/
}