Problem:
Score : 500 points

Problem Statement

There is a building with n rooms, numbered 1 to n.

We can move from any room to any other room in the building.

Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).

Initially, there was one person in each room in the building.

After that, we know that there were exactly k moves happened up to now.

We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?

Find the count modulo (10^9 + 7).

Constraints

All values in input are integers.

3 \leq n \leq 2 \times 10^5

2 \leq k \leq 10^9

Input

Input is given from Standard Input in the following format:

n k

Output

Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).

Sample Input 1

3 2

Sample Output 1

10

Let c_1, c_2, and c_3 be the number of people in Room 1, 2, and 3 now, respectively. There are 10 possible combination of (c_1, c_2, c_3):

(0, 0, 3)

(0, 1, 2)

(0, 2, 1)

(0, 3, 0)

(1, 0, 2)

(1, 1, 1)

(1, 2, 0)

(2, 0, 1)

(2, 1, 0)

(3, 0, 0)

For example, (c_1, c_2, c_3) will be (0, 1, 2) if the person in Room 1 goes to Room 2 and then one of the persons in Room 2 goes to Room 3.

Sample Input 2

200000 1000000000

Sample Output 2

607923868

Print the count modulo (10^9 + 7).

Sample Input 3

15 6

Sample Output 3

22583772

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*by freesteed*/
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
#define fill(x,c) memset(x,c,sizeof(x))
#define fi first
#define se second
#define sz(A) (int)(A.size())
#define all(x) (x).begin(),(x).end()
#define caset int ___T; cin>>___T; for(int cs=1;cs<=___T;cs++)
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<int> VI;
const ll mod=1e9+7;
const ll oo=1e9+10;
const int Max = 1e6+10;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int ri() {
	int n;
	scanf("%d", &n);
	return n;
}
int64_t rs64() {
	int64_t n;
	scanf("%" SCNd64, &n);
	return n;
}
#define MOD 1000000007
template<int mod>
struct ModInt{
	int x;
	ModInt():x(0){}
	ModInt(long long y):x(y>=0?y%mod:(mod-(-y)%mod)%mod){}
	ModInt &operator+=(const ModInt &p){
		if((x+=p.x)>=mod)x-=mod;
		return *this;
	}
	ModInt &operator-=(const ModInt &p){
		if((x+=mod-p.x)>=mod)x-=mod;
		return *this;
	}
	ModInt &operator*=(const ModInt &p){
		x=(int)(1LL*x*p.x%mod);
		return *this;
	}
	ModInt &operator/=(const ModInt &p){
		*this*=p.inverse();
		return *this;
	}
	ModInt &operator^=(long long p){
		ModInt res = 1;
		for (; p; p >>= 1) {
			if (p & 1) res *= *this;
			*this *= *this;
		}
		return *this = res;
	}
	ModInt operator-()const{return ModInt(-x);}
	ModInt operator+(const ModInt &p)const{return ModInt(*this)+=p;}
	ModInt operator-(const ModInt &p)const{return ModInt(*this)-=p;}
	ModInt operator*(const ModInt &p)const{return ModInt(*this)*=p;}
	ModInt operator/(const ModInt &p)const{return ModInt(*this)/=p;}
	ModInt operator^(long long p)const{return ModInt(*this)^=p;}
	bool operator==(const ModInt &p)const{return x==p.x;}
	bool operator!=(const ModInt &p)const{return x!=p.x;}
	explicit operator int() const { return x; }						   // added by QCFium
	ModInt operator=(const int p) {x = p; return ModInt(*this);} // added by QCFium
	ModInt inverse()const{
		int a=x,b=mod,u=1,v=0,t;
		while(b>0){
			t=a/b;
			a-=t*b;
			std::swap(a,b);
			u-=t*v;
			std::swap(u,v);
		}
		return ModInt(u);
	}
	friend std::ostream &operator<<(std::ostream &os,const ModInt<mod> &p){
		return os<<p.x;
	}
	friend std::istream &operator>>(std::istream &is,ModInt<mod> &a){
		long long x;
		is>>x;
		a=ModInt<mod>(x);
		return (is);
	}
};
typedef ModInt<MOD> mint;

struct MComb {
	std::vector<mint> fact;
	std::vector<mint> inversed;
	MComb(int n) { // O(n+log(mod))
		fact = std::vector<mint>(n+1,1);
		for (int i = 1; i <= n; i++) fact[i] = fact[i-1]*mint(i);
		inversed = std::vector<mint>(n+1);
		inversed[n] = fact[n] ^ (MOD-2);
		for (int i = n - 1; i >= 0; i--) inversed[i]=inversed[i+1]*mint(i+1);
	}
	mint ncr(int n, int r) {
		return (fact[n] * inversed[r] * inversed[n-r]);
	}
	mint npr(int n, int r) {
		return (fact[n] * inversed[n-r]);
	}
	mint nhr(int n, int r) {
		assert(n+r-1 < (int)fact.size());
		return ncr(n+r-1, r);
	}
};
//mint ncr(int n, int r) {
//	mint res = 1;
//	for (int i = n - r + 1; i <= n; i++) res *= i;
//	for (int i = 1; i <= r; i++) res /= i;
//	return res;
//}
int help() {
	int n = ri(), k = ri();
	MComb com(2 * n + 1);
	mint res = 0;
	rep(m,0,n){
		if (m > k) break;
		res += com.ncr(n, m) * com.nhr(n-m,m);
	}
	std::cout << res << std::endl;
	return 0;
}
//ll a[Max];
void solve(){
	help(); 
}
int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	//caset{
		solve();
	//}
  	return 0;
}