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 <iostream>
#include <vector>
using namespace std;
typedef long long ll;
#define MOD 1000000007LL

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  ll n, k; cin >> n >> k;

  auto pow = [](ll a, int n) {
    ll res = 1;
    for (; n; n >>= 1) {
      if (n & 1) res = res * a % MOD;
      a = a * a % MOD;
    }
    return res;
  };

  vector<ll> table(n+1), invTable(n+1);
  table[0] = 1;
  for (int i = 1; i <= n; i++) {
    table[i] = table[i-1] * i % MOD;
  }
  invTable[n] = pow(table[n], MOD-2);
  for (int i = n; i > 0; i--) {
    invTable[i-1] = invTable[i] * i % MOD;
  }

  auto comb = [&](int n, int r) {
    ll tmp = invTable[n-r] * invTable[r] % MOD;
    return table[n] * tmp % MOD;
  };

  ll ans = 0;
  for (int i = 0; i <= min(k, n-1); i++)  {
    ans += comb(n, i) * comb(n-1, i) % MOD;
    ans %= MOD;
  }
  cout << ans << "\n";
}