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 <string>
#include <algorithm>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <deque>
#include <set>
#include <iomanip>
#include <utility>
typedef long long ll;
typedef long double ld;
using namespace std;

ll mod=1000000007;
const int SIZE=510000;
ll fac[SIZE], finv[SIZE], inv[SIZE];
void COMinit(ll m){
    fac[0]=fac[1]=1;
    finv[0]=finv[1]=1;
    inv[1]=1;
    for(int i=2; i<SIZE; ++i){
        fac[i]=(fac[i-1]*i)%m;
        inv[i]=m-(inv[m%i]*(m/i))%m;
        finv[i]=(finv[i-1]*inv[i])%m;
    }
}
ll COM(int n, int k, ll m){
    if(n<k) return 0;
    if(n<0 || k<0) return 0;
    return (fac[n]*((finv[k]*finv[n-k])%m))%m;
}

int main() {
	int n, k;
	cin >> n >> k;
	COMinit(mod);
	ll ans=0;
	for(int i=0; i<=min(n-1, k); ++i){
		ll t=(COM(n, i, mod)*COM(n-1, i, mod))%mod;
		ans=(ans+t)%mod;
	}
	cout << ans << endl;
	return 0;
}