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 <limits.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <math.h>

int comp(const void *a, const void *b){return *(int*)a-*(int*)b;}

#define P 1000000007

long m(long a, int x){
	if(x==0) return 1;
	if(x%2==0){
		return m(a*a%P, x/2);
	}else{
		return m(a*a%P, x/2)*a%P;
	}
}

int r(int a){
	return m(a, P-2);
}

int kh[400001];

long c(int a, int b){
	// kh[a] / kh[b] / kh[a-b];
	return ((long)kh[a]) * r(((long)kh[b]) * kh[a-b] %P) %P;
}

int main(void){
	int n, k;
	scanf("%d%d", &n, &k);

	kh[0]=kh[1]=1;
	for(long i=1; i<=n*2; i++){
		kh[i+1] = kh[i] * (i+1) %P;
	}
	long ans=0;
	if(k>=n){
		ans=c(n*2-1, n-1);
	}else{
		for(int i=0; i<=k; i++){
			ans += c(n, i) * c(n-1, i) % P;
		}
	}
	ans %= P;
	printf("%d\n", ans);
}