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<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+5;
const ll mod=1e9+7;
ll n,k;
ll power(ll a,ll b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}
ll F[N],I[N];
void init(){
    F[0]=1;
    for(ll i=1;i<=N;i++){
        F[i]=F[i-1]*i%mod;
    }
    I[N]=power(F[N],mod-2);
    for(ll i=N;i--;){
        I[i]=I[i+1]*(i+1)%mod;
    }
}

ll C(ll n, ll k){
    return F[n]*I[n-k]%mod*I[k]%mod;
}

int main(){
    init();
    cin>>n>>k;
    ll sum=0;
    for(ll i=0;i<n&&i<=k;i++){
       sum=(sum+C(n,i)*C(n-1,i))%mod;
    }
    cout<<sum;
    return 0;
}