// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define fi first

#define se second

#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)

#define endl '\n'

#define ld long double

typedef pair<ll,ll> pii;

const int maxn=200005;

const int mod=1e9+7;

ll qp(ll a,ll b)

{

    ll ret=1;

    while(b)

    {

        if(b&1) ret=ret*a%mod;

        b>>=1;

        a=a*a%mod;

    }

    return ret;

}

ll jc[maxn],inv[maxn];

void init()

{

    jc[0]=1;

    for(int i=1;i<=200004;i++) jc[i]=jc[i-1]*i%mod;

    inv[200004]=qp(jc[200004],mod-2);

    for(int i=200003;i>=0;i--) inv[i]=inv[i+1]*(i+1)%mod;

}

ll C(int x,int y)

{

    return jc[x]*inv[x-y]%mod*inv[y]%mod;

}

ll calc(int x,int y)

{

    return C(x+y-2,x-1);

}

int main()

{

    #ifdef local

    freopen("in.txt","r",stdin);

    #endif // local

    IOS;

    init();

    int h,w,a,b;

    cin>>h>>w>>a>>b;

    ll ans=0;

    for(int i=h-a+1;i<=h;i++)

    {

        ans+=calc(i,b)*calc(h-i+1,w-b)%mod;

        ans%=mod;

    }

    ans=calc(h,w)-ans;

    ans=(ans+mod)%mod;

    cout<<ans<<endl;

}


// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

#pragma warning (disable:4996)



using namespace std;

using ll = long long;

using LL = ll;

using ull = unsigned long long;

using uLL = ull;

#define clean_0(A) memset((A),0,sizeof(A))

#define clean_1(A) memset((A),-1,sizeof(A))

#define For(i,a,b) for(int (i)=(a);(i)<=b;++(i))

#define Max_(a,b) ((a)>=(b)?(a):(b))

#define Min_(a,b) ((a)<=(b)?(a):(b))

constexpr auto Max = [](const auto& a, const auto& b) {return a >= b ? a : b; };

constexpr auto Min = [](const auto& a, const auto& b) {return a <= b ? a : b; };

constexpr int INF = 0x7F7F7F7F;

constexpr double eps = 1e-9;



constexpr ll mod1 = 197;

constexpr ll mod2 = 19260817;

constexpr ll mod3 = ll(1e8) + 7;

constexpr ll mod4 = ll(1e9) + 7;;

constexpr ll mod5 = 998244353;

constexpr ll mod = mod4;



constexpr int maxn = int(3e5) + 1;

constexpr int MAXN = maxn;

namespace io {



#define in(a) (a=read())

#define out(a) write(a)

#define outn(a) (out(a),putchar('\n'))



#define I_int  int

	inline I_int read() {

		I_int x = 0, f = 1; char c = getchar();

		while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }

		while (c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }

		return x * f;

	}

	template<typename T>

	bool scan_d(T& ret)

	{

#define p(x) ((x)>='0'&&(x)<='9')

		int c; int sgn;

		if (c = getchar(), c == EOF)return false;

		while (c != '-' && !p(c))c = getchar();

		sgn = (c == '-') ? -1 : 1;

		ret = (c == '-') ? 0 : c - '0';

		while (c = getchar(), p(c))ret = ret * 10 + c - '0';

		ret *= sgn;

		return true;

#undef p

	}

	char F[200];

	template<typename T>

	inline void write(T x) {

		if (x == 0) { putchar('0'); return; }

		auto tmp = x > 0 ? x : -x;

		if (x < 0) putchar('-');

		int cnt = 0;

		while (tmp > 0) {

			F[cnt++] = tmp % 10 + '0';

			tmp /= 10;

		}

		while (cnt > 0) putchar(F[--cnt]);

	}

	void read(char* s, int Arr[], size_t n)

	{

		for (size_t i = 0; i < n; ++i)Arr[i] = strtoll(s, &s, 10);

	}

#undef I_int

}

using namespace io;



template<typename T>

T exgcd(T a, T b, T& x, T& y)

{

	if (a == 0 && b == 0)return -1;

	if (b == 0) { x = 1, y = 0; return a; }

	T d = exgcd(b, a % b, y, x);

	y -= a / b * x;

	return d;

}

template<typename T>

T inv1(T a, T Mod = mod)

{

	T x, y;

	return exgcd(a, Mod, x, y) == 1 ? (x % Mod + Mod) % Mod : -1;

}

template<typename T>

constexpr T _gcd(T a, T b)

{

	return b ? _gcd(b, a % b) : a;

}

template<typename T>

constexpr T combinator(T n, T m, T Mod = mod - 1)

{

	m = min(m, n - m);

	T ret = 1;

	for (T i = 1; i <= m; ++i)

	{

		ret = ret * (n - m + i) / i % Mod;

	}

	return ret;

}

template<typename T>

constexpr T poww(T a, T b, T Mod = mod)

{

	T ret = 1;

	a %= Mod;

	b %= Mod - 1;

	if (b < 0)b += Mod - 1;

	while (b)

	{

		if (b & 1)ret = ret * a % Mod;

		a = a * a % Mod;

		b >>= 1;

	}

	return ret;

}

constexpr auto inv2 = [](ll a, ll Mod = mod) {return poww(a, Mod - 2, Mod); };

struct Combinator

{

	static constexpr int Mod = ::mod;

	static constexpr int Maxn = 1;//::maxn;

	int dp[Maxn][Maxn];//this is C(n,m)%(mod-1)

	constexpr Combinator() :dp()

	{

		for (int i = 0; i < Maxn; ++i)

		{

			for (int j = 0; j <= i; ++j)

			{

				if (j == 0) dp[i][j] = 1;

				else if (i == j) dp[i][j] = 1;

				else dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1]) % (Mod - 1);

			}

		}

	}

};

ll Fact[maxn], Inv[maxn];

//void Init(int n)

//{

//	Fact[0] = 1;

//	for (int i = 1; i <= n; i++)

//		Fact[i] = Fact[i - 1] * i % mod;

//	Inv[0] = 1;

//	Inv[n] = poww(Fact[n], mod - 2LL);

//	for (int i = n - 1; i > 0; i--)

//		Inv[i] = Inv[i + 1] * (i + 1LL) % mod;

//}

//ll C(int a, int b) {

//	if (a < 0 || b < 0)return 1;

//	return F[a] * Inv[b] % mod * Inv[a - b] % mod;

//}

//int main()

//{

//	Init(2.1e5);

//	int a, b, h, w;

//	while (cin >> h>>w>>a>>b)

//	{

//		ll ans = C(h + w - 2, h - 1);

//		for (ll i = 1; i <= b; i++) {

//			ans -= C(h - a + i - 2, h - a - 1) * C(a - 1 + w - i, a - 1) % mod;

//			ans = (ans % mod + mod) % mod;

//		}

//		printf("%lld", ans);

//	}

//}

int H, W, A, B;

const ll Maxn = maxn;

const int Mod = mod;

void Init(int N)

{

	Fact[0] = 1;

	for (int i = 1; i <= N; ++i)

		Fact[i] = Fact[i - 1] * i % Mod;

	Inv[0] = 1;

	Inv[N] = inv2(Fact[N], Mod);//poww(Fact[N], Mod - 2LL);

	for (int i = N; i > 1; --i)

		Inv[i - 1] = Inv[i] * i % Mod;

}

ll C(int a, int b) {

	if (a < 0 || b < 0)return 1;

	return Fact[a] * Inv[b] % Mod * Inv[a - b] % Mod;

}

int main() {

	scanf("%d %d %d %d", &H, &W, &A, &B);

	Init(H + W - 2);

	ll ans = C(H + W - 2, H - 1);

	for (ll i = 1; i <= B; i++) {

		ans -= C(H - A + i - 2, H - A - 1) * C(A - 1 + W - i, A - 1) % Mod;

		ans = (ans % Mod + Mod) % Mod;

	}

	printf("%lld", ans);

}

// Fastest known solution for this problem: