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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int k = scanner.nextInt();
        System.out.println(roaming(n,k));
    }
    public static int roaming(int n, int k){
        int m = 1_000_000_007;
        BigInteger bigM = BigInteger.valueOf(m);
        if(k>=n-1){
            long factB = 1;
            for (int i = 2; i <= n-1; i++) {
                factB = factB*i%m;
            }
            long topF = 1;
            for (int i = 2*n-1; i >= n+1; i--) {
                topF = topF*i%m;
            }
            return BigInteger.valueOf(topF).multiply(BigInteger.valueOf(factB).modInverse(bigM)).mod(bigM).intValue();
        }else{
            BigInteger s = BigInteger.ONE;
            BigInteger tmp = BigInteger.ONE;
            for (int i = 1; i <= k; i++) {
                tmp = tmp.multiply(BigInteger.valueOf(n-i)).mod(bigM).multiply(BigInteger.valueOf(n-i+1))
                        .multiply(BigInteger.valueOf(i).modInverse(bigM).pow(2));
                s = s.add(tmp).mod(bigM);
            }
            return s.intValue();
        }
    }
}