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: Accepted)

#include <stdio.h>

int mod = 1000000007;

long long mpow(long long a, int n)
{
	long long res = 1;
	while (n > 0) {
		if(n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

int minv(int a)
{
	return mpow(a, mod-2);
}

int main(void)
{
	int n, k;
	long long mfact[200005];
	long long nPr[200005];
	
	scanf("%d%d", &n, &k);
	
	if (n <= k) k = n - 1;
	
	mfact[0] = 1; nPr[0] = 1;
	for (int i = 1; i <= 200004
	; i++) {
		mfact[i] = mfact[i-1] * i % mod;
		nPr[i] = nPr[i-1] * (n-i+1) % mod;
	}
	
	long long ans = 1;
	
	for (int i = 1; i <= k; i++)
		ans = (ans + (nPr[i] * minv(mfact[i]) % mod) * (nPr[i+1] * minv(n) % mod * minv(mfact[i]) % mod) % mod) % mod;
	
	printf("%lld", ans);
	
	return 0;
}