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

const int MAX = 510000;
const int MOD = 1000000007;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0]=fac[1]=1;
    finv[0]=finv[1]=1;
    inv[1]=1;
    for(int i=2;i<MAX;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;
    }
}
long long COM(int n,int k){
    if(n<k) return 0;
    if(n<0||k<0) return 0;
    return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;
}

int main(){
  COMinit();
  ll n,k;
  cin>>n>>k;
  k=min(n-1,k);
  ll ans=0;
  for(int i=0;i<=k;i++){
    ans+=(COM(n-i-1+i,i)%MOD)*(COM(n,i)%MOD);
    ans%=MOD;
  }
  cout<<ans<<endl;
}
### Prompt

In cpp, your task is to solve the following problem:

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).

Examples

Input

3 2


Output

10


Input

200000 1000000000


Output

607923868


Input

15 6


Output

22583772

### Response

```cpp
#include<bits/stdc++.h>
#define ll long long
using namespace std;

const int MAX = 510000;
const int MOD = 1000000007;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0]=fac[1]=1;
    finv[0]=finv[1]=1;
    inv[1]=1;
    for(int i=2;i<MAX;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;
    }
}
long long COM(int n,int k){
    if(n<k) return 0;
    if(n<0||k<0) return 0;
    return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;
}

int main(){
  COMinit();
  ll n,k;
  cin>>n>>k;
  k=min(n-1,k);
  ll ans=0;
  for(int i=0;i<=k;i++){
    ans+=(COM(n-i-1+i,i)%MOD)*(COM(n,i)%MOD);
    ans%=MOD;
  }
  cout<<ans<<endl;
}
```