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)

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include "bits/stdc++.h" // define macro "/D__MAI"

using namespace std;
typedef long long int ll;

#define xprintf(fmt,...) fprintf(stderr,fmt,__VA_ARGS__)
#define debugv(v) {printf("L%d %s => ",__LINE__,#v);for(auto e:v){cout<<e<<" ";}cout<<endl;}
#define debugm(m) {printf("L%d %s is..\n",__LINE__,#m);for(auto v:m){for(auto e:v){cout<<e<<" ";}cout<<endl;}}
#define debuga(m,w) {printf("L%d %s is => ",__LINE__,#m);for(int x=0;x<(w);x++){cout<<(m)[x]<<" ";}cout<<endl;}
#define debugaa(m,w,h) {printf("L%d %s is..\n",__LINE__,#m);for(int y=0;y<(h);y++){for(int x=0;x<(w);x++){cout<<(m)[x][y]<<" ";}cout<<endl;}}
#define debugaar(m,w,h) {printf("L%d %s is..\n",__LINE__,#m);for(int y=0;y<(h);y++){for(int x=0;x<(w);x++){cout<<(m)[y][x]<<" ";}cout<<endl;}}
#define ALL(v) (v).begin(),(v).end()
#define repeat(cnt,l) for(auto cnt=0;cnt<(l);++cnt)
#define iterate(cnt,b,e) for(auto cnt=(b);cnt!=(e);++cnt)
#define MD 1000000007ll
#define PI 3.1415926535897932384626433832795
template<typename T1, typename T2>
ostream& operator <<(ostream &o, const pair<T1, T2> p) { o << "(" << p.first << ":" << p.second << ")"; return o; }

mt19937 mt(8901016);
inline int rand_int(int l, int h) {
    return uniform_int_distribution<>(l, h)(mt);
}

#ifdef __MAI
#define getchar_unlocked getchar
#define putchar_unlocked putchar
#endif
#ifdef __VSCC
#define getchar_unlocked _getchar_nolock
#define putchar_unlocked _putchar_nolock
#endif
namespace {
#define isvisiablechar(c) (0x21<=(c)&&(c)<=0x7E)
    class MaiScanner {
    public:
        template<typename T> void input_integer(T& var) {
            var = 0;
            T sign = 1;
            int cc = getchar_unlocked();
            for (; cc<'0' || '9'<cc; cc = getchar_unlocked())
                if (cc == '-') sign = -1;
            for (; '0' <= cc&&cc <= '9'; cc = getchar_unlocked())
                var = (var << 3) + (var << 1) + cc - '0';
            var = var*sign;
        }
        inline int c() { return getchar_unlocked(); }
        inline MaiScanner& operator>>(int& var) {
            input_integer<int>(var);
            return *this;
        }
        inline MaiScanner& operator>>(long long& var) {
            input_integer<long long>(var);
            return *this;
        }
        inline MaiScanner& operator>>(string& var) {
            int cc = getchar_unlocked();
            for (; !isvisiablechar(cc); cc = getchar_unlocked());
            for (; isvisiablechar(cc); cc = getchar_unlocked())
                var.push_back(cc);
        }
        template<typename IT> void in(IT begin, IT end) {
            for (auto it = begin; it != end; ++it) *this >> *it;
        }
    };
}
MaiScanner scanner;

template<typename T> inline T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template<typename T> inline T lcm(T a, T b) { return a * b / gcd(a, b); }

template<typename iterator> inline size_t argmin(iterator begin, iterator end) {
    return distance(begin, min_element(begin, end));
}
template<typename iterator> inline size_t argmax(iterator begin, iterator end) {
    return distance(begin, max_element(begin, end));
}

template<typename iterator, typename T, typename F>
T reduce(iterator begin, iterator end, T init, F func) {
    T sum = init;
    for (; begin != end; ++begin)
        if (!func(sum, *begin)) break;
    return sum;
}


class llmod {
public: const ll MOD = MD;
private:
    ll val;
    inline ll cut(ll v) const { return ((v%MOD) + MOD) % MOD; }
public:

    llmod() : MOD(MD), val(0) {}
    llmod(ll num, ll m = MD) : MOD(m), val(cut(num)) {}
    llmod(const llmod& lm, ll m) : MOD(m), val(lm.val) {}

    inline ll operator*() const { return val; }
    inline llmod& operator=(const llmod& lm) { val = lm.val; return *this; }
    inline llmod& operator=(ll v) { val = cut(v); return *this; }

    inline llmod& operator+=(ll v) { val = cut(val + v); return *this; }
    inline llmod& operator+=(const llmod& l) { val = cut(val + l.val); return *this; }
    inline llmod& operator-=(ll v) { val = cut(val - v); return *this; }
    inline llmod& operator-=(const llmod& l) { val = cut(val - l.val); return *this; }
    inline llmod& operator*=(ll v) { val = cut(val * v); return *this; }
    inline llmod& operator*=(const llmod& l) { val = cut(val * l.val); return *this; }
    inline llmod& operator++() { val = (val + 1) % MOD; return *this; }
    inline llmod operator++(int) { llmod t = *this; val = (val + 1) % MOD; return t; }
};
ostream& operator<<(ostream& os, const llmod& l) { os << *l; return os; }

inline llmod operator+(llmod t, const llmod& r) { return t += r; }
inline llmod operator-(llmod t, const llmod& r) { return t -= r; }
inline llmod operator*(llmod t, const llmod& r) { return t *= r; }



// MEMO : 逆元...powm(n,MD-2)
llmod pow(llmod x, ll p) {
    llmod y = 1;
    while (0 < p) {
        if (p % 2)
            y *= x;
        x *= x;
        p /= 2;
    }
    return y;
}

inline llmod& operator/=(llmod& l, const llmod& r) { return l *= pow(r, r.MOD - 2); }


ll gen[100010];
llmod genfact[100010];
llmod fact[200010];

ll height, width, aa, bb;

void gengen() {
    gen[0] = 1;
    for (ll i = 1; i < 100005; ++i) {
        gen[i] = *pow(llmod(i), MD - 2);
    }
    genfact[0] = 1;
    genfact[1] = gen[1];
    for (ll x = 2; x <= 100010; ++x) {
        genfact[x] = gen[x] * genfact[x - 1];
    }
    fact[0] = 1;
    fact[1] = 1;
    for (ll x = 2; x <= 200010; ++x) {
        fact[x] = x * fact[x - 1];
    }
}

llmod movecount(ll n,ll m) {
    --n; --m;
    if (n == 0 || m == 0) return llmod(1);
    llmod result = 1;

    // for (ll x = n + 1; x <= n + m; ++x) {
    //     result *= x;
    // }
    // for (ll x = 1; x <= m; ++x) {
    //     result *= gen[x];
    // }
    return fact[n+m]*genfact[n]*genfact[m];
}

int main() {

    cin >> height >> width >> aa >> bb;


    gengen();

    llmod result = 0;

    for (ll y = 0; y < height - aa; ++y) {
        llmod r = movecount(y + 1, bb);
        r *= movecount(height - y, width - bb);
        result += r;
    }

    cout << *result << endl;

    return 0;
}