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)

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
#define chmin(a,b) a=min(a,b)
#define chmax(a,b) a=max(a,b)
#define rep(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
#define mad(a,b) a=(a+b)%mod
#define N 500050
ll po(ll x,ll y){
	ll res=1;
	for(;y;y>>=1){
		if(y&1)res=res*x%mod;
		x=x*x%mod;
	}
	return res;
}
ll inv[N],fac[N],ivf[N];
ll C(ll n,ll k){
	return fac[n]*ivf[n-k]%mod*ivf[k]%mod;
}
ll H(ll n,ll k){
	return C(n+k-1,k);
}
int main(){
	for(ll i=1;i<N;i++)inv[i]=po(i,mod-2);
	fac[0]=ivf[0]=1;
	for(ll i=1;i<N;i++){
		fac[i]=fac[i-1]*i%mod;
		ivf[i]=ivf[i-1]*inv[i]%mod;
	}
	ll n,k;
	cin>>n>>k;
	ll ans=0;
	for(ll x=0;x<=min(k,n-1);x++){
		ans=(ans+H(n-x,x)*C(n,x))%mod;
	}
	cout<<ans<<endl;
}