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: Wrong Answer)

#include <bits/stdc++.h>
#define MOD 1000000007
typedef long long ll;

using namespace std;

ll power(ll a, ll n) {
    if (n == 0) {
        return 1LL;
    }
    ll half = power(a, n / 2) % MOD;
    ll p = (half * half) % MOD;
    if (n % 2 != 0) {
        return (p * a) % MOD;
    } else {
        return p;
    }
}

ll comb(ll n, ll r) {
    ll ans = 1;
    for (int i = 0; i < r; i++) {
        ans = ans * power(i + 1, MOD - 2) % MOD;
        ans = ans * (n - i) % MOD;
    }
    return ans;
}


int main() {
    ll n, k, ans;
    ans = 0LL;
    cin >> n >> k;
    if (n - 1 <= k) {
        ans = comb(2 * n - 1, n - 1);
    } else {
        // n C m
        vector<ll> values(k + 1);
        for (int m = 0; m <= k; m++) {
            if (m == 0) {
                values[m] = 1;
            } else {
                values[m] = (n - m + 1) * power(m, MOD - 2) * values[m - 1];
                values[m] %= MOD;
            }
        }
        // for (int m = 0; m <= k; m++) {
        //     auto v = values[m];
        //     cout << v << endl;
        //     cout << v * ((n - m) * power(n, MOD - 2)) % MOD << endl;
        // }
        for (int m = 0; m <= k; m++) {
            // cout << comb(n, m) << ", " << comb(n - 1, m) << endl;
            // ans += comb(n, m) * comb(n - 1, m);
            ll v = values[m] * (n - m) * power(n, MOD - 2) % MOD;
            ans += values[m] * v;
            ans %= MOD;
        }
    }
    cout << ans << endl;
    return 0;
}