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>
using namespace std;
typedef long long ll;
const int N = 2e5 + 87;
const ll MOD = 1e9 + 7;
ll f[N], fi[N];
ll qpow(ll a, ll b)
{
    ll r = 1;
    for (;b;b>>=1,a=a*a%MOD)
        if (b&1)
            r = r*a%MOD;
    return r;
}
ll c(ll n, ll k)
{
    return f[n] * fi[k] % MOD * fi[n - k] % MOD;
}
ll h(ll n, ll k)
{
    return c(n + k - 1, k);
}
int main()
{
    f[0] = f[1] = fi[0] = fi[1] = 1;
    for (int i = 2; i < N; ++i)
        f[i] = i * f[i - 1] % MOD;
    for (int i = 2; i < N; ++i)
        fi[i] = (MOD - MOD / i) * fi[MOD % i] % MOD;
    for (int i = 2; i < N; ++i)
        fi[i] = fi[i] * fi[i - 1] % MOD;
    ll n, k;
    cin >> n >> k;
    ll ans = 0;
    for (int i = 0; i <= min(n - 1, k); ++i)
        (ans += h(n - i, i) * c(n, i)) %= MOD;
    cout << ans << endl;
}