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>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
using ll = long long;

constexpr int P = 1000000007;

ll powmod(ll n, ll k) {
    ll r = 1, t = n % P;
    for (; k != 0; k /= 2) {
        if (k & 1) r = r * t % P;
        t = t * t % P;
    }
    return r;
}

ll inv(int n) {
    int m = P, y = 0, v = 1;
    while (1) {
        int q = m / n;
        int r = m % n;
        if (r == 0) return v < 0 ? v + P : v;
        y -= v * q;
        swap(y, v);
        m = n;
        n = r;
    }
}

int comb2(int n, int k) {
    if (n < k || k < 0) return 0;
    ll t = 1;
    k = min(k, n - k);
    for (int i = 2; i <= k; i++) {
        t = (t * i) % P;
    }
    t = inv(t);
    for (int i = n - k + 1; i <= n; i++) {
        t = (t * i) % P;
    }
    return t;
}

vector<int> f0, f1;
void init(int n) {
    f0.resize(n + 1);
    f0[0] = 1;
    for (int i = 1; i <= n; i++) {
        f0[i] = (ll)f0[i - 1] * i % P;
    }
    f1.resize(n + 1);
    f1[n] = inv(f0[n]);
    for (int i = n; i > 0; i--) {
        f1[i - 1] = (ll)f1[i] * i % P;
    }
}
ll fact(int k) {
    return f0[k];
}
ll comb(int n, int k) {
    if (n < k || k < 0) return 0;
    return (ll)f0[n] * f1[k] % P * f1[n - k] % P;
}

int main() {
    int n, k;
    cin >> n >> k;

    if (k < n - 1) {
        init(n);
        ll r = 0;
        for (int i = 0; i <= k; i++) {
            int m = n - i - 1;
            r += comb(n - 1, m) * comb(n, i) % P;
        }
        cout << r % P << endl;

    } else {
        cout << comb2(n + (n - 1), n) << endl;
    }

    //N-1回で全部できる 1回のときは場合分け なかった
    //2回以上N-2回以下はそれ以下でできたものはできる 0の個数がK個以下

    return 0;
}