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>
#include <stdlib.h>
#define P (1000000007) //10^9+7
#define N (200010)

long mfac(long,long);		// a! mod p
long mpow(long,long,long);	// a^b mod p
long mcom(long,long,long);	// aCb mod p

long fact[N]={1};
long invfact[N]={1};

int main(){
	long n,k;
	long count=0;
	scanf("%ld %ld",&n,&k);
	k = k<n?k:n;
	for(int i=1;i<=n;i++){
		fact[i] = fact[i-1]*i%P;
		invfact[i] = mpow(fact[i],P-2,P);
	}
	for(int i=0;i<=k;i++){
		count += mcom(n,i,P)*mcom(n-1,i,P)%P;
		count %= P;
	}
	printf("%ld",count);
	return 0;
}

long mfac(long a,long p){
	if(a==0) return 1;
	return a*mfac(a-1,p)%p;
}

long mpow(long a,long b,long p){
	long n=1;
	while(b){
		if(b&1) n = n*a%p;
		b = b>>1;
		a = a*a%p;
	}
	return n;
}

long mcom(long a,long b,long p){
	long n = 1;
	n = n*fact[a]%p;
	n = n*invfact[a-b]%p;
	n = n*invfact[b]%p;
	return n;
}