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 (1.42.0) (judge verdict: Accepted)

use std::io;
use std::str::FromStr;

static MOD: u64 = 1_000_000_007;

fn input_tuple() -> (u64, u64) {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    let vec: Vec<u64> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
    (vec[0], vec[1])
}

fn binary_power(x: u64, n: u64) -> u64 {
    if n == 0 {
      return 1;
    }
    if n % 2 == 0 {
      return (binary_power(x, n / 2) % MOD).pow(2) % MOD;
    } else {
      return binary_power(x, n - 1) % MOD * x % MOD;
    }
}
  
fn inverse_mod(n: u64) -> u64 {
    binary_power(n, MOD-2)
}  

fn combination(n: u64, mut m: u64) -> u64 {
    if n < m {
        return 0;
    }
    
    if m > n - m {
        m = n - m
    }

    let mut ret = 1u64;
    for i in 0..m {
        ret %= MOD;
        ret *= n - i;        
    }
    ret %= MOD;

    let mut div = 1u64;
    for i in 1..(m+1) {
        div %= MOD;
        div *= i;
    }
    div %= MOD;

    // println!("{}, {}", ret, div);
    ret * inverse_mod(div) % MOD
}

fn main() {
    let (n, k) = input_tuple();
    if k >= n {
        // 全組み合わせ
        println!("{}", combination(n * 2 - 1, n - 1));
        return;
    }
    let mut sum = 0;
    let mut c1 = combination(n, n - k);
    let mut c2 = combination(n - 1, k);
    for i in (n-k)..(n+1) {
        // println!("{}, {}", c1, c2);
        sum = sum + c1 % MOD * c2 % MOD;
        sum %= MOD;

        c1 = c1 * (n - i) % MOD * inverse_mod(i + 1) % MOD;
        c2 = c2 * (n - i) % MOD * inverse_mod(i) % MOD;
    }
    println!("{}", sum);
}