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 rep(i,n) for(int i=0;i<n;i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> pp;
const int INF = 1e9;
const int MOD = 1000000007;

const int MAX = 1000000;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
        fac[0] = fac[1] = 1;
        finv[0] = finv[1] = 1;
        inv[1] = 1;
        for (int i = 2; i < MAX; i++){
                fac[i] = fac[i - 1] * i % MOD;
                inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
                finv[i] = finv[i - 1] * inv[i] % MOD;
        }
}

// 二項係数計算
long long COM(int n, int k){
        if (n < k) return 0;
        if (n < 0 || k < 0) return 0;
        return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}



int main() {
        int n;
        ll k;
        ll ans = 0;
        cin >> n >> k;
        if(k ==1){
                ans = (n*(n-1))%MOD;
        }
        else if(k < n){
                COMinit();
                rep(i,k+1){
                           ans = ( ans + (COM(n-1,i) * COM(n,i))%MOD)%MOD;
                }

        }else{
          COMinit();
          rep(i,n){
            ans = ( ans + (COM(n-1,i) * COM(n,i))%MOD)%MOD;
          }
        }

        cout << ans << endl;
        return 0;
}