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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#define rep(i, n) for (int i = 0; i < (int) (n); i++)
#define MOD 1000000007
typedef long long ll;
ll PMRS(ll N, ll P, ll M){
	if (P == 0){
		return 1;
	}else if (P % 2 == 0){
		ll t = PMRS(N, P/2, M); 
		return t * t % M;
	}else{
		return N * PMRS(N, P - 1, M) % M;
	}
}  

ll inv_Felmat(ll a, ll M){
	return PMRS(a, M - 2, M);
}

ll perm(ll n, ll k, ll M){
	if (k == 1) return n;
	return perm(n, k - 1, M) * (n - k + 1) % M;
}

ll comb(ll n, ll k, ll M){
	ll x = perm(n, k, M);
	ll y = perm(k, k, M);
	return x * inv_Felmat(y, M);
}

int main(){
	ll n, k, ans = 0;
	scanf("%d%d",&n,&k);
	if (n > k) k = n - 1;
	rep(i, k)	ans = (ans + comb(n, i, MOD) *comb(n - 1, i, MOD)) % MOD;
	printf("%d%d", ans);
}