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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

long long int M=1e+9+7;
long long int power(long long int x, long long int n)
{
    if (n == 0)
        return (long long)1;
    else if (n % 2 == 0) //n is even
        return power((x * x) % M, n / 2);
    else //n is odd
        return (x * power((x * x) % M, (n - 1) / 2)) % M;
}
long long int cans(long long int p, long long int q)
{
    return ((p * power(q, M - 2)) % M + M) % M;
}
const int N=4e+5+5;
long long int n,fact[N],facti[N];

int main()
{
    long long int a;
    cin>>n>>a;
    // cout<<a<<" "<<b<<" ";
    fact[0]=facti[0]=1;
    for(int i=1;i<N;i++)
    {
        fact[i]=(i*fact[i-1])%M;
        facti[i]=cans(1,fact[i]);
    }
    long long int ans=fact[2*n-1];
    ans=(ans*facti[n])%M;
    ans=(ans*facti[n-1])%M;
    cout<<ans;
}