//  Created by conan1024hao in 2020.
//  Copyright © 2020 conan1024hao. All rights reserved.
//  専用ライブラリです、自由にコピーして構いません。
//  感谢看我的代码！Wechat:conan1024hao Twitter/QQ:810396815
#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define IINF 10e8
#define INF 10e17
#define MOD 1000000007
#define mod 1000000007
#define rep(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
//using Int=int_fast64_t;
typedef long long int ll;
typedef long long int LL;
typedef pair<ll,ll>Pll;
typedef pair<int,int>Pin;
template<class T> using V=vector<T>;
long long GCD(long long a, long long b) { return b ? GCD(b, a%b) : a; }
long long LCM(long long a, long long b)  {return a/GCD(a,b)*b;}
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int ddx[8]={-1,0,1,0,1,1,-1,-1};
int ddy[8]={0,-1,0,1,1,-1,1,-1};
char dir[4]={'u','l','d','r'};
ll cmp1(pair<ll,string>a,pair<ll,string> b){
        if(a.fi!=b.fi)
        return a.fi<b.fi;
        else
        return a.se<b.se;
}

//---------------------------------------------------------------------------
struct mint{
    using ull=unsigned long long int;
    ull v;
    mint(ll vv=0){s(vv%mod+mod);}
    mint& s(ull vv){
        v=vv<mod?vv:vv-mod;
        return *this;
    }
    //オーバーロード
    mint operator-()const{return mint()-*this;}//符号反転
    mint&operator+=(const mint&val){return s(v+val.v);}
    mint&operator-=(const mint&val){return s(v+mod-val.v);}
    mint&operator*=(const mint&val){
        v=ull(v)*val.v%mod;
        return *this;
    }
    mint&operator/=(const mint&val){return *this*=val.inv();}
    mint operator+(const mint&val){return mint(*this)+=val;}
    mint operator-(const mint&val){return mint(*this)-=val;}
    mint operator*(const mint&val){return mint(*this)*=val;}
    mint operator/(const mint&val){return mint(*this)/=val;}
    mint pow(int n)const{
        mint res(1),x(*this);
        while(n){
            if(n&1)res*=x;
            x*=x;
            n>>=1;
        }
        return res;
    }
    mint inv()const{return pow(mod-2);}
    //拡張ユークリッドの互除法
    /* mint inv()const{
        int x,y;
        int g=extgcd(v,mod,x,y);
        assert(g==1);
        if(x<0)x+=mod;
        return mint(x);
    }*/
    friend ostream& operator<<(ostream&os,const mint&val){
        return os<<val.v;
    }//出力
    bool operator<(const mint&val)const{return v<val.v;}
    bool operator==(const mint&val)const{return v==val.v;}
    bool operator>(const mint&val)const{return v>val.v;}
};
const ll MAX = 2000010;//設定
mint fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void init(){
    fac[0] = fac[1] = 1;
    for(int i=1;i<MAX;i++)fac[i]=fac[i-1]*i;
    finv[MAX-1]=fac[MAX-1].inv();
    for(int i=MAX-2;i>=0;i--)finv[i]=finv[i+1]*(i+1);
    for(int i=MAX-2;i>=1;i--)inv[i]=finv[i]+fac[i-1];
}
//階乗
mint factor(ll n,ll k){
  if (n<k) return 0;
  if (n<0 || k<0) return 0;
  return fac[n]*finv[k];
}
// 二項係数計算
mint 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];
}
//---------------------------------------------------------------------------
int main(){//問題をちゃんと見ろ!!!!!!!!!!!!!!!!!      llか??????????统一类型        memset()!!!!!!!!!!!!       ペナを減らせ!!!!!!!!!!!!!
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------
    ll n,k;cin>>n>>k;
    init();
    mint ans=0;
    for(ll i=0;i<=mmin(k,n-1);i++){
        ans+=COM(n,i)*COM(n-1,i);
    }
    cout<<ans<<endl;
    //-------------------------------
    return 0;
}
//---------------------------------------------------------------------------





Measured memory_bytes: 47104