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;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}


template<typename T,T MOD = 1000000007>
struct Mint{
  static constexpr T mod = MOD;
  T v;
  Mint():v(0){}
  Mint(signed v):v(v){}
  Mint(long long t){v=t%MOD;if(v<0) v+=MOD;}

  Mint pow(long long k){
    Mint res(1),tmp(v);
    while(k){
      if(k&1) res*=tmp;
      tmp*=tmp;
      k>>=1;
    }
    return res;
  }

  static Mint add_identity(){return Mint(0);}
  static Mint mul_identity(){return Mint(1);}

  Mint inv(){return pow(MOD-2);}

  Mint& operator+=(Mint a){v+=a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator-=(Mint a){v+=MOD-a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator*=(Mint a){v=1LL*v*a.v%MOD;return *this;}
  Mint& operator/=(Mint a){return (*this)*=a.inv();}

  Mint operator+(Mint a) const{return Mint(v)+=a;}
  Mint operator-(Mint a) const{return Mint(v)-=a;}
  Mint operator*(Mint a) const{return Mint(v)*=a;}
  Mint operator/(Mint a) const{return Mint(v)/=a;}

  Mint operator-() const{return v?Mint(MOD-v):Mint(v);}

  bool operator==(const Mint a)const{return v==a.v;}
  bool operator!=(const Mint a)const{return v!=a.v;}
  bool operator <(const Mint a)const{return v <a.v;}

  static Mint comb(long long n,int k){
    Mint num(1),dom(1);
    for(int i=0;i<k;i++){
      num*=Mint(n-i);
      dom*=Mint(i+1);
    }
    return num/dom;
  }
};
template<typename T,T MOD> constexpr T Mint<T, MOD>::mod;
template<typename T,T MOD>
ostream& operator<<(ostream &os,Mint<T, MOD> m){os<<m.v;return os;}


template<typename M>
class Enumeration{
private:
  static vector<M> fact,finv,invs;
public:
  static void init(int n){
    n=min<decltype(M::mod)>(n,M::mod-1);

    int m=fact.size();
    if(n<m) return;

    fact.resize(n+1,1);
    finv.resize(n+1,1);
    invs.resize(n+1,1);

    if(m==0) m=1;
    for(int i=m;i<=n;i++) fact[i]=fact[i-1]*M(i);
    finv[n]=M(1)/fact[n];
    for(int i=n;i>=m;i--) finv[i-1]=finv[i]*M(i);
    for(int i=m;i<=n;i++) invs[i]=finv[i]*fact[i-1];
  }

  static M Fact(int n){
    init(n);
    return fact[n];
  }
  static M Finv(int n){
    init(n);
    return finv[n];
  }
  static M Invs(int n){
    init(n);
    return invs[n];
  }

  static M C(int n,int k){
    if(n<k||k<0) return M(0);
    init(n);
    return fact[n]*finv[n-k]*finv[k];
  }

  static M P(int n,int k){
    if(n<k||k<0) return M(0);
    init(n);
    return fact[n]*finv[n-k];
  }

  // put n identical balls into k distinct boxes
  static M H(int n,int k){
    if(n<0||k<0) return M(0);
    if(!n&&!k) return M(1);
    init(n+k);
    return C(n+k-1,n);
  }
};
template<typename M>
vector<M> Enumeration<M>::fact=vector<M>();
template<typename M>
vector<M> Enumeration<M>::finv=vector<M>();
template<typename M>
vector<M> Enumeration<M>::invs=vector<M>();

//INSERT ABOVE HERE
signed main(){
  using M = Mint<int>;
  using E = Enumeration<M>;
  E::init(1e6);

  int n,k;
  cin>>n>>k;
  M ans{0};
  for(int i=0;i<=min(n,k);i++)
    ans+=E::C(n,i)*E::H(i,n-i);

  cout<<ans<<endl;
  return 0;
}