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: Wrong Answer)

use std::io::{self, BufRead};

const M: i64 = 1000000007;

fn pow(p: i64, n: i64) -> i64 {
  let mut q = n;
  let mut m = p;
  let mut ans = 1;
  while q > 0 {
    if q % 2 == 1 {
      ans = (ans * m) % M;
    }
    q /= 2;
    m = (m * m) % M;
  }
  ans
}

fn div(n: i64, m: i64) -> i64 {
  (n * pow(m, M - 2)) % M
}

fn mul(n: i64, m: i64) -> i64 {
  (n * m) % M
}

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let line = lines.next().unwrap().unwrap();
  let vs: Vec<i64> = line.split(' ').map(|x| x.parse().unwrap()).collect();
  let n = vs[0];
  let k = vs[1];
  let mut facts = vec![1 as i64; (2 * n) as usize];

  for i in 2..(2*n) {
    facts[i as usize] = mul(facts[(i-1) as usize], i);
  }

  let ans = if n < k {
    div(facts[(2 * n - 1) as usize], mul(facts[(n - 1) as usize], facts[n as usize]))
  } else {
    (0..(k+1)).map(|m| div(mul(facts[n as usize], facts[(n - 1) as usize]), mul(mul(mul(facts[m as usize], facts[(n - m) as usize]), facts[if n > m { (n - m - 1) as usize } else { 1 }]), facts[m as usize]))).fold(0, |acc, x| (acc + x) % M)
  };

  println!("{}", ans);
}