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>
#define int long long 
using namespace std;
const int MODN=1e9+7;
const int MAXN=2e5+10;
int fact[MAXN];
int quick_pow(int a,int b,int modn=MODN){
    a%=MODN;
    int ret=1;
    // int cheng=a;
    while(b){
        if(b&1)ret*=a;
        ret%=MODN;
        a*=a;
        a%=MODN;
        b>>=1;
    }
    return ret;
}
int C(int n,int r){
    return fact[n]*(quick_pow((fact[n-r]*fact[r]),MODN-2))%MODN;
}

int32_t main(){
    
    int n,k;cin>>n>>k;
    fact[0]=1;
    for(int i=1;i<MAXN;i++)fact[i]=(fact[i-1]*i)%MODN;
    int sum=0;
    for(int i=0;i<=min(k,n-1);i++){
        sum+=C(n,i)*C(n-1,i);
        sum%=MODN;
    }
    cout<<sum<<endl;
	return 0;
}