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)

//ヘッダー
#include<bits/stdc++.h>
using namespace std;

//型定義
typedef long long ll;

//定数

const ll MOD=1e+9+7; //10^9+7
const int SIZE=2e+5+1;
//REPマクロ
#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define REPD(i,n) for(ll i=n-1;i>=0;i--)
#define REP2(i,a,b) for(ll i=a;i<=(ll)(b);i++)
#define REPD2(i,a,b) for(ll i=a;i>=(ll)(b);i--)

//グローバル変数


//ライブラリ挿入
vector<long long> fact, fact_inv, inv;
 
/*  init_nCk :二項係数のための前処理
    計算量:O(n)
*/
void init_nCk(int SIZE) {
    fact.resize(SIZE + 5);
    fact_inv.resize(SIZE + 5);
    inv.resize(SIZE + 5);
    fact[0] = fact[1] = 1;
    fact_inv[0] = fact_inv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < SIZE + 5; i++) {
        fact[i] = fact[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        fact_inv[i] = fact_inv[i - 1] * inv[i] % MOD;
    }
}
 
/*  nCk :MODでの二項係数を求める(前処理 int_nCk が必要)
    計算量:O(1)
*/
long long nCk(int n, int k) {
    assert(!(n < k));
    assert(!(n < 0 || k < 0));
    return fact[n] * (fact_inv[k] * fact_inv[n - k] % MOD) % MOD;
}


int main(){
    ll N,k;
    cin >> N >> k;
    ll ans=0;
    init_nCk(SIZE);
    REP(i,min(N,k+1)){
        ans+=((nCk(N,i)*nCk(N-1,i))%MOD);
        ans%=MOD;
    }
    
    cout << ans << endl;

}