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).

Examples

Input

3 2


Output

10


Input

200000 1000000000


Output

607923868


Input

15 6


Output

22583772
import java.util.Scanner;

/**
 * https://www.hamayanhamayan.com/entry/2020/02/24/092051
 */
public class Main {
    private static final int MOD = 1_000_000_007;
    private static final String YES = "Yes";
    private static final String NO = "No";

    private static long[] fact;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int k = sc.nextInt();

        System.out.println(solve(n, k));
    }

    private static long solve(int n, int k) {
        long count = 0;
        fact = factInit(n, MOD);

        for (int numZeroRoom=0; numZeroRoom<=Math.min(n, k); numZeroRoom++) {
            // System.err.println("\t" + (n-numZeroRoom) + "P" + (numZeroRoom) + " = " + permutation(n-numZeroRoom,  numZeroRoom, MOD));
            count += combination(n, numZeroRoom, MOD) * combination(n-1, n - (n-numZeroRoom), MOD) % MOD;
            count %= MOD;
            // System.err.println("zero(" + numZeroRoom + "): +" + tmp + " => " + count);
        }

        return count;
    }

    public static long combination(int n, int r, long p) {
        if (n == 0 && r == 0) return 1;
        if (n < r || n < 0)   return 0;

        return (fact[n] * power((fact[r]), p-2, p) % p) * power(fact[n-r], p-2, p) % p;
    }

    public static long[] factInit(int n, long p) {
        long[] fact = new long[n+1];
        fact[0] = 1;
        fact[1] = 1;
        for (int i=2; i<=n; i++) {
            fact[i] = i * fact[i-1] % p;
        }

        return fact;
    }


    public static long power(long base, long t, long p) {
        long ans = 1;

        while (t > 0) {
            if (t%2 == 1) {
                ans = ans * base % p;
            }

            t /= 2;
            base = base * base % p;
        }

        return ans;
    }
}