#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pii pair<int, int>
#define pli pair<ll, int>
#define pil pair<int, ll>
#define pll pair<ll, ll>
#define eb emplace_back
#define all(v) v.begin(), v.end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep3(i, l, n) for (int i = l; i < (n); ++i)
#define sz(v) (int)v.size()
#define endl '\n'
const int inf = 1000000007;
const ll INF = 1e18;
// int mod = 998244353;
int mod = 1000000007;
const double eps = 1e-8;
#define abs(x) (x >= 0 ? x : -(x))
#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())
#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b); }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
template<typename T, typename U> T pow_(T a, U b) { return b ? pow_(a * a, b / 2) * (b % 2 ? a : 1) : 1; }
ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }
template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }
template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { rep(i, sz(vec)) { if (i) os << " "; os << vec[i]; } return os; }
template<class T> ostream& operator << (ostream& os, const set<T>& s) { for (auto si : s) os << si << " "; return os; }
template<typename T> inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }
template<class T, class T2> inline void add(T &a, T2 b) { (a += (b % mod + mod)) %= mod; }
template<class T> void operator += (vector<T>& v, vector<T> v2) { rep(i, sz(v2)) v.eb(v2[i]); }

void solve();

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(12);
    int T;
    T = 1;
    while (T--) solve();
}

struct mint {
    ll x;
    mint (ll x = 0) : x((mod + x % mod) % mod) {}
    mint operator -() const { return mint(-x); }
    mint& operator += (const mint a) { // コンストラクタの、modで割った余りがくる
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator -= (const mint a) {
        if ((x += -a.x + mod) >= mod) x -= mod;
        return *this;
    }
    mint& operator *= (const mint a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator + (const mint a) const {
        mint res(*this);
        return res += a; // ?
    }
    mint operator - (const mint a) const {
        mint res(*this);
        return res -= a;
    }
    mint operator * (const mint a) const {
        mint res(*this);
        return res *= a;
    }
    mint pow(ll t) const { // mint(a).pow(t) tは余りとったりしたらあかん
        if (t == 0) return 1;
        mint a = pow(t >> 1); // 再帰
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }
    // 逆元
    mint inv() const {
        return pow(mod - 2);
    }
    mint& operator /= (const mint a) { // invを利用
        return (*this) *= a.inv();
    }
    mint operator / (const mint a) const {
        mint res(*this);
        return res /= a;
    }
    friend ostream &operator << (ostream &s, mint a) { return s << a.x; }
};

struct COM {
    vector<mint> fact, ifact;
    COM(int n) : fact(n + 1), ifact(n + 1) {
        assert(n < mod);
        fact[0] = 1;
        rep3(i, 1, n + 1) 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]; // n! / (k! * (n-k)!)
    }
} com(2000000);

// https://atcoder.jp/contests/abc156/submissions/10323391
void solve() {
    int n, k;
    cin >> n >> k;
    
    if (k >= n - 1) {
        cout << com(n + (n - 1), n - 1) << endl;
        return;
    }
    
    // 最初0の個数0
    // k回の移動で最大k個0にできる
    mint ans = 0;
    rep(i, k + 1) { // 0の個数
        // n-i個非0
        // n個の○とn-i-1個の仕切り, ただし0個の場所なし
        int x = n - (n - i); // 先に各場所に1つ配る
        ans += com(x + (n - i - 1), x) * com(n, i); // 0の場所の選び方も忘れずに
    }
    cout << ans << endl;
}


Measured memory_bytes: 34528