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)

#define mod 1000000007
char buf[30];sz;
char out[10];osz=10;
int N,K;
long I[2<<17]={1,1};
main()
{
	read(0,buf,30);
	for(;;)
	{
		char c=buf[sz++];
		if(c<'0')break;
		N=N*10+c-'0';
	}
	for(;;)
	{
		char c=buf[sz++];
		if(c<'0')break;
		K=K*10+c-'0';
	}
	for(int i=2;i<=K+1;i++)I[i]=(mod-mod/i)*I[mod%i]%mod;
	long ans=0;
	long x=1,y=1;
	for(long i=0;i<N&&i<=K;i++)
	{
		ans+=x*y%mod;
		x=x*(N-i-1)%mod*I[i+1]%mod;
		y=y*(N-i)%mod*I[i+1]%mod;
	}
	ans%=mod;
	do{
		out[--osz]=ans%10+'0';
	}while(ans/=10);
	write(1,out+osz,10-osz);
}