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: Time Limit Exceeded)

// https://atcoder.jp/contests/abc156/tasks/abc156_e
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
using namespace std;

// 繰り返し二乗法を用いて a の n 乗を算出する(ただし mod m の空間内)
long long modpow(long long a, long long n, long long mod)
{
    long long ret = 1;
    while (n > 0)
    {
        if (n & 1)
        {
            ret *= a;
            ret = ret % mod;
        }
        a *= a;
        a = a % mod;
        n = (n >> 1);
    }
    return ret;
}

long long modnCk(long long n, long long k, long long mod)
{
    if (k == 0) return 1;
    long long ret = 1, x = 1, y = 1;
    for (long long i = 0; i < k; ++i)
    {
        x *= (n - i);
        x = x % mod;
 
        y *= (k - i);
        y = y % mod;
    }
    ret = modpow(y, mod - 2, mod) % mod;
    ret = (ret * x) % mod;
    return ret;
}

int main()
{
    long long n, k;
    cin >> n >> k;
    static constexpr long long mod = 1e+9 + 7;

    // 空いている部屋の数に関してループを回す
    long long ans = 0;
    if (k >= n - 1)
    {
        ans += modnCk(2 * n - 1, n, mod);
        ans = ans % mod;
        if (ans < 0) ans = (ans + mod) % mod;
        cout << ans << endl;
        return 0;
    }
    ans += 1;
    for (long long i = 1; i <= k; ++i)
    {
        long long cnt = modnCk(n, i, mod) * modnCk(n - 1, i, mod);
        cnt %= mod;
        if (cnt < 0)
            cnt = (cnt + mod) % mod;
        ans += cnt;
        ans = ans % mod;
        if (ans < 0)
            ans = (ans + mod) % mod;
    }

    cout << ans << endl;
    return 0;
}