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) throws Exception {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		sc.close();

		int mod = 1000000007;
		NCR ncr = new NCR(n * 2, mod);

		if (k >= n - 1) {
			System.out.println(ncr.calc(n + n - 1, n));
		} else {
			long ans = 0;
			for (int i = 0; i <= k; i++) {
				long val0 = ncr.calc(n, i);
				long val1 = ncr.calc(n - 1, i);;
				ans += val0 * val1;
				ans %= mod;
			}
			System.out.println(ans);
		}
	}

	static class NCR {
		long[] p, pi;
		int m;

		public NCR(int n, int mod) {
			n++;
			m = mod;
			p = new long[n];
			pi = new long[n];
			p[0] = 1;
			pi[0] = 1;
			for (int i = 1; i < n; i++) {
				p[i] = p[i - 1] * i % m;
			}
			pi[n - 1] = BigInteger.valueOf(p[n - 1])
					.modInverse(BigInteger.valueOf(m)).longValue();
			for (int i = n - 1; i > 1; i--) {
				pi[i - 1] = pi[i] * i % m;
			}
		}

		public long calc(int n, int r) {
			if (n < r) return 0;
			return p[n] * pi[r] % m * pi[n - r] % m;
		}
	}
}