#include <bits/stdc++.h>
#define forn(i,s,t) for(int i = (s); i <= (t); ++i)
#define form(i,s,t) for(int i = (s); i >= (t); --i)
#define rep(i,s,t) for(int i = (s); i < (t); ++i)
using namespace std;
const int N = 2e5 + 5, Mod = 1e9 + 7;
namespace Modint {
	struct Mint {
		int res;
		Mint() {}
		Mint(int _r) : res(_r) {}
		inline friend Mint operator + (const Mint& A, const Mint& B) {
			return Mint((A.res + B.res >= Mod) ? (A.res + B.res - Mod) : (A.res + B.res));
		}
		inline friend Mint operator - (const Mint& A, const Mint& B) {return A + Mint(Mod - B.res); }
		inline friend Mint operator * (const Mint& A, const Mint& B) {return Mint(1ll * A.res * B.res % Mod); }
		inline friend Mint& operator += (Mint& A, const Mint& B) {return A = A + B; }
		inline friend Mint& operator -= (Mint& A, const Mint& B) {return A = A - B; }
		inline friend Mint& operator *= (Mint& A, const Mint& B) {return A = A * B; }
		inline friend Mint q_pow(Mint p, int k = Mod - 2) {
			Mint res(1);
			for (; k; k >>= 1, p *= p) (k & 1) && (res *= p, 0);
			return res;
		}
	} ;
}
using Modint :: Mint;
Mint fac[N], ifac[N];
inline void table(int n) {
	fac[0] = Mint(1);
	forn (i, 1, n) fac[i] = fac[i - 1] * Mint(i);
	ifac[n] = q_pow(fac[n]);
	form (i, n - 1, 0) ifac[i] = ifac[i + 1] * Mint(i + 1);
}
inline bool check(int n, int r) {return n - r >= 0 && n >= 0 && r >= 0; }
inline Mint C(int n, int r) {
	if (!check(n, r)) return Mint(0);
	return fac[n] * ifac[r] * ifac[n - r];
}
int n, k;
int main() {
	scanf("%d%d", &n, &k), table(n);
	Mint res(0), tmp(1); int ll = max(k - n, 0), rr = max(k - 1, 0) - 1;
	int l = ll, r = ll - 1;
//	printf("[%d, %d]\n", l, r);
	while (r <= rr) ++r, (r && (tmp *= Mint(r), 0));
	while (l <= ll) (l && (tmp *= q_pow(Mint(l)), 0)), ++l;
//		printf("[%d, %d] %d\n", l, r, tmp.res);
	forn (a, 1, n) {
		ll = a + k - n, rr = a + k - 1;
//		if (a == 1) printf("%d %d\n", ll, rr);
		if (check(a + k - 1, n)) {
			while (r < rr) ++r, tmp *= Mint(r);
			while (l < ll) tmp *= q_pow(Mint(l)), ++l;
		} else continue ;
//		printf("[%d, %d] %d %d %d\n", l, r, tmp.res, C(n - 1, a - 1).res, ifac[n].res);
//		printf("%d\n", (C(n - 1, a - 1) * tmp * ifac[n]).res);
		res += C(n - 1, a - 1) * tmp * ifac[n];
	}
	printf("%d\n", res.res);
	return 0;
}