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 Rust (1.15.1) (judge verdict: Time Limit Exceeded)

fn main() {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    println!("{}", solve(&input));
}

const MODULO: i64 = 1_000_000_007;

fn solve(input: &str) -> i64 {
    let mut iter = input.split_whitespace().map(|w| w.parse().unwrap());
    let n: i64 = iter.next().unwrap();
    let k: i64 = iter.next().unwrap();
    if k >= n - 1 {
        binom(2 * n - 1, n - 1)
    } else {
        let mut ans = 0;
        for m in 0..k + 1 {
            ans = (ans + binom(n - 1, n - m - 1) * binom(n, m) % MODULO) % MODULO;
        }
        ans
    }
}

fn power(base: i64, exponent: i64) -> i64 {
    let mut acc = 1;
    let mut base = base;
    let mut exponent = exponent;
    while exponent != 0 {
        if exponent & 1 == 1 {
            acc = acc * base % MODULO;
        }
        base = base * base % MODULO;
        exponent >>= 1;
    }
    acc
}

fn inverse(x: i64) -> i64 {
    power(x, MODULO - 2)
}

fn binom(n: i64, k: i64) -> i64 {
    let mut numer = 1;
    let mut denom = 1;
    for i in 0..k {
        numer = numer * (n - i) % MODULO;
        denom = denom * (k - i) % MODULO;
    }
    numer * inverse(denom) % MODULO
}