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 <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<bool> vb;
typedef vector<string> vst;
typedef pair<ll,ll> pll;
typedef vector<pll> vpll;

const ll mod=1e9+7;
const ll inf=1e16; 

#define rep(i,m,n) for(ll i=m;i<n;i++)
#define repr(i,m,n) for(ll i=m;i>=n;i--)

const long nmax=1000000;


long long fac[nmax],finv[nmax],inv[nmax];

void comint(){
 fac[0]=fac[1]=1;
 finv[0]=finv[1]=1;
 inv[1]=1;
 for(long i=2;i<nmax;i++){
  fac[i]=fac[i-1]*i%mod;
  inv[i]=mod-inv[mod%i]*(mod/i)%mod;
  finv[i]=finv[i-1]*inv[i]%mod;
}
}

auto com(auto n, auto k){
 if(n<k||n<0||k<0) return 0;
 return fac[n]*(finv[k]*finv[n-k]%mod)%mod;
}




int main(){
   cin.tie(nullptr);
  ios::sync_with_stdio(false);

  comint();
  
  ll n,k;
  cin >> n >> k;
  ll res=0;
  rep(m,0,min(n,k)+1) res+=(com(n,m)*com(n-1,n-m-1))%mod,res%=mod;
   
 cout << res << endl;
}