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: Accepted)

#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
template<class Read>void in(Read &x){
    x=0;
    int f=0;
    char ch=getchar();
    while(ch<'0'||ch>'9'){
        f|=(ch=='-');
        ch=getchar();
    }
    while(ch>='0'&&ch<='9'){
        x=(x<<1)+(x<<3)+(ch^48);
        ch=getchar();
    }
    x=f?-x:x;
    return;
}
const int M=1e9+7;
int n,m;
ll ans=1,f[1000005]={1},g[1000005]={1};
ll ksm(ll x,ll y){
    ll s=1;
    while(y){
        if(y&1)s=(s*x)%M;
        x=(x*x)%M;
        y>>=1;
    }
    return s;
}
ll work(int x,int y){
    return f[x]%M*g[y]%M*g[x-y]%M;
}
int main(){
	in(n);in(m);
    for(int i=1;i<=n;i++){
        f[i]=f[i-1]*i%M;
        g[i]=ksm(f[i],M-2);
    }
    int k=min(n-1,m);
    for(int i=1;i<=k;i++){
        ll x=work(n,i);
        ll y=work(n-1,i);
        ans=(ans+x*y%M)%M;
    }
    cout<<ans<<endl;
	return 0;
}