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 <vector>
using namespace std;
long long const mod = 1000000007;
long long mod_fact(int x){
    static vector<long long> fact = {1, 1};
    if(x < 0)return 0; //
    if(x < fact.size())return fact[x];
    for(int i = fact.size(); i <= x; i++){
        fact.push_back((fact[i-1] * i) % mod);
    }
    return fact[x];
}

long long mod_Montmort_number(int x){
    static vector<long long> mont = {1, 0};
    if(x < 0)return 0; //
    if(x < mont.size())return mont[x];
    for(int i = mont.size(); i <= x; i++){
        mont.push_back(((i-1) * (mont[i-1] + mont[i-2])) % mod);
        //mont.push_back((i * mont[i-1] + ((i%2 == 0) ? 1 : -1)) % mod);
    }
    return mont[x];
}

long long mod_pow(long long a, long long b){
    long long res = 1;
    a %= mod;
    while(b > 0){
        if(b%2 == 1) res = (res * a) % mod;
        a = (a * a) % mod;
        b /= 2;
    }
    return res;
}

long long mod_inv(long long a){
    return mod_pow(a, mod-2);
}

long long mod_comb(int a, int b){
    if(b < 0 || a < b)return 0;
    if(b > a/2)return mod_comb(a, a-b);
    return (((mod_fact(a) * mod_inv(mod_fact(a-b))) % mod) * mod_inv(mod_fact(b))) % mod;
}
/*
long long mod_comb(long long a, long long b){
    if(b < 0 || a < b)return 0;
    if(b > a/2)return mod_comb(a, a-b);
    long long res = 1;
    for(long long i = 0; i < b; i++){
        res = (res * ((a-i)%mod)) % mod;
        res = (res * mod_inv(b-i)) % mod;
    }
    return res;
}
*/
int main(void){
    int n, k;
    cin >> n >> k;
    if(k == 1){
        // XD
    }
    k = min(k, n-1);
    long long ans = 0;
    for(int z=0;z<=k;z++){
        long long t = mod_comb(n, z) * mod_comb(n-1, z);
        ans += t%mod;
    }
    cout << ans%mod << endl;
}