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>

#define rep(i, n) for (ll i = 0; i < (n); i++)
#define rep2(i, a, b) for (ll i = (a); i < (b); i++)
#define chmax(x, v)                                                            \
  do {                                                                         \
    x = max(x, v);                                                             \
  } while (0)
#define chmin(x, v)                                                            \
  do {                                                                         \
    x = min(x, v);                                                             \
  } while (0)
typedef uint64_t ull;
typedef int64_t ll;
typedef std::pair<ll, ll> PLL;

using namespace std;

std::vector<ll> fac(300000, 1), finv(300000, 1), inv(300000, 1);

void c4_init(ll m) {
    for (ll i = 2; i < 300000; i++) {
        fac[i] = fac[i - 1] * i % m;
        inv[i] = m - inv[m%i] * (m / i) % m;
        finv[i] = finv[i - 1] * inv[i] % m;
    }
}

ll c4(ll n, ll k, ll m) {
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % m) % m;
}

// {{{ Modint
// using mint = modint<1000000007>;
// mint x = 2;
// std::cout << x.a << std::endl;
template <std::int64_t M>
class modint {
 public:
  int64_t a;

  explicit constexpr modint(const int64_t x = 0) noexcept :
    a((x % M + M) % M) {}
  constexpr int64_t &value() noexcept { return a; }
  constexpr const int64_t &value() const noexcept { return a; }
  constexpr modint operator+(const modint rhs) const noexcept {
    return modint(*this) += rhs;
  }
  constexpr modint operator-(const modint rhs) const noexcept {
    return modint(*this) -= rhs;
  }
  constexpr modint operator*(const modint rhs) const noexcept {
    return modint(*this) *= rhs;
  }
  constexpr modint operator/(const modint rhs) const noexcept {
    return modint(*this) /= rhs;
  }
  constexpr modint &operator+=(const modint rhs) noexcept {
    a += rhs.a;
    if (a >= M) {
      a -= M;
    }
    return *this;
  }
  constexpr modint &operator-=(const modint rhs) noexcept {
    if (a < rhs.a) {
      a += M;
    }
    a -= rhs.a;
    return *this;
  }
  constexpr modint &operator*=(const modint rhs) noexcept {
    a = a * rhs.a % M;
    return *this;
  }
  constexpr modint &operator/=(modint rhs) noexcept {
    int64_t exp = M - 2;
    while (exp) {
      if (exp % 2) {
        *this *= rhs;
      }
      rhs *= rhs;
      exp /= 2;
    }
    return *this;
  }
  constexpr modint pow(ll t) const {
    if (!t) return modint(1);
    modint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
};
// }}}

const ll M = 1e9+7;
using mint = modint<M>;
signed main() {
  ll n,k;
  cin >> n >> k;
  c4_init(M);
  mint ans(0);
  for (ll t = max((ll)1, n-k); t <= n; t++) {
    ans += mint(c4(n,t,M)) * mint(c4(n-1,t-1,M));
  }
  cout<<ans.value()<<endl;
  return 0;
}