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++ (GCC 9.2.1) (judge verdict: Accepted)

/*#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>

#include <boost/rational.hpp>
*/
#include <bits/stdc++.h>
using namespace std;
#define _GLIBCXX_DEBUG
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
using ll = long long;
using ld = long double;
#define rep2(i, s, n) for (ll i = (s); i < (ll)(n); i++)
#define rep3(i, n) for (ll i = 1; i < (ll)(n+1); i++)
#define rep4(i, s, n) for (ll i = (s); i < (ll)(n+1); i++)
#define repr(i,n) for (ll i = (n-1); i>=0;i--)
#define repr3(i,n) for(ll i = (n);i>0;i--)
#define repr4(i,s,n) for(ll i = (n);i>=(s);i--)
#define stlen(s) ll s.size()-1
#define all(v) v.begin(), v.end()
#define allr(v) v.rbegin(), v.rend()
#define cout(n) cout<<std::fixed<<std::setprecision(n)
using Graph = vector<vector<int>>;
using Graphw = vector<vector<pair<ll,ll>>>;
#define INF1  INT_MAX;
#define INF2  LLONG_MAX;
#define PI  3.14159265358979323846;
#define MOD 1000000007;
/*
namespace mp = boost::multiprecision;
// 任意長整数型
using Bint = mp::cpp_int;
// 仮数部長が32の浮動小数点数型
using Real32 = mp::number<mp::cpp_dec_float<32>>;
// 仮数部長が1024の浮動小数点数型
using Real1024 = mp::number<mp::cpp_dec_float<1024>>;
// 有理数型
using Rat = boost::rational<Bint>;
*/
const int mod = 1000000007;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {//階乗、その逆元も存在する。
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
  mint p(int n,int k){
  return fact[n]*ifact[n-k];
  }
}comb(1000005);//100005まで計算（どこまで必要かで変更）

int main(){
int n;
ll k;
cin>>n>>k;
if(n<=k){
    cout<<comb(2*n-1,n)<<endl;
}
else{
    mint ans=0;
    rep(i,k+1){
        ans+=comb(n,i)*(comb(n-1,i));
    }
    cout<<ans<<endl;
}
}