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;

#ifdef LOCAL
#include "dump.hpp"
#else
#define dump(...)
#define dumpv(...)
#endif

#define rep(i, n) for (int i = 0; i < (n); i++)
#define mins(x, y) (x = min(x, y))
#define maxs(x, y) (x = max(x, y))
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vvi = vector<vi>;
using vvl = vector<vl>;
using P = pair<int, int>;
const int MOD = 1e9 + 7;
const int INF = 1001001001;
const ll LINF = 1001002003004005006ll;

// struct mint {
//   ll v;
//   mint() : v(0) {}
//   mint(ll v) : v((v % MOD + MOD) % MOD) {}
//   operator ll() const { return v; }
//   mint pow(ll t) const;
//   mint inv() const;
// };
// mint& operator+=(mint& a, mint b) { return a = a.v + b.v; }
// mint& operator-=(mint& a, mint b) { return a = a.v - b.v; }
// mint& operator*=(mint& a, mint b) { return a = a.v * b.v; }
// mint operator+(mint a, mint b) { return a += b; }
// mint operator-(mint a, mint b) { return a -= b; }
// mint operator*(mint a, mint b) { return a *= b; }
// mint operator-(mint a) { return (mint)0 - a; }
// mint& operator/=(mint& a, mint b) { return a *= b.inv(); }
// mint operator/(mint a, mint b) { return a /= b; }
// mint mint::pow(ll t) const {
//   if (!t) return 1;
//   mint a = pow(t >> 1);
//   a *= a;
//   if (t & 1) a *= *this;
//   return a;
// }
// mint mint::inv() const { return pow(MOD - 2); }
// bool operator==(mint a, mint b) { return a.v == b.v; }
// bool operator!=(mint a, mint b) { return a.v != b.v; }
// istream& operator>>(istream& os, mint& a) { return (os >> a.v); }
// ostream& operator<<(ostream& os, const mint& a) { return (os << a.v); }

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 {
    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 {
    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 {
    mint res(*this);
    return res/=a;
  }
};

mint mfact(int n) { // n! (mod MOD)
  static vector<mint> fact(1, 1);
  int pn = fact.size();
  if (pn < n + 1) {
    fact.resize(n + 1);
    for (int i = pn; i <= n; i++) fact[i] = fact[i - 1] * (mint)i;
  }
  return fact[n];
}
mint mifact(int n) { // n! (mod MOD)
  static vector<mint> ifact(1, 1);
  int pn = ifact.size();
  if (pn < n + 1) {
    ifact.resize(n + 1);
    ifact[n] = mfact(n).inv();
    for (int i = n; i >= pn; --i) ifact[i - 1] = ifact[i] * (mint)i;
  }
  return ifact[n];
}
mint nPk(int n, int k) {
  if (n < 0 || k < 0 || n < k) return 0;
  if (k == 0) return 1;
  return mfact(n) * mfact(n - k).inv();
}
mint nCk(int n, int k) {
  if (n < 0 || k < 0 || n < k) return 0;
  if (k == 0) return 1;
  // return mfact(n) * mfact(k).inv() * mfact(n - k).inv();
  return mfact(n) * mifact(k) * mifact(n - k);
}

void solve() {
  int n, k;
  cin >> n >> k;
  k = min(k, n - 1);
  mint ans = 0;
  mifact(n);
  for (int i = 0; i <= k; i++) {
    ans += nCk(n, i) * nCk(n - i - 1 + i, i);
  }
  cout << ans.x << endl;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
  // freopen("temp.1", "r", stdin);
  solve();
  return 0;
}