Levko loves permutations very much. A permutation of length n is a sequence of distinct positive integers, each is at most n.

Let’s assume that value gcd(a, b) shows the greatest common divisor of numbers a and b. Levko assumes that element pi of permutation p1, p2, ... , pn is good if gcd(i, pi) > 1. Levko considers a permutation beautiful, if it has exactly k good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them.

Input

The single line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ n).

Output

In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.

If there are multiple suitable permutations, you are allowed to print any of them.

Examples

Input

4 2


Output

2 4 3 1

Input

1 1


Output

-1

Note

In the first sample elements 4 and 3 are good because gcd(2, 4) = 2 > 1 and gcd(3, 3) = 3 > 1. Elements 2 and 1 are not good because gcd(1, 2) = 1 and gcd(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.

The second sample has no beautiful permutations.
import java.io.BufferedInputStream;
import java.util.Scanner;

/**
 * Created by jizhe on 2015/12/30.
 */
public class LevkoAndPermutation {
    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        StringBuilder out = new StringBuilder();

        int N = in.nextInt();
        int K = in.nextInt();

        if( K == N )
        {
            System.out.printf("-1\n");
            return;
        }
        for( int i = 0; i < N-K-1; i++ )
        {
            out.append(i+2+" ");
        }
        out.append(1+" ");
        for( int i = N-K; i < N; i++ )
        {
            out.append(i+1+" ");
        }
        System.out.printf("%s\n", out.toString());
    }
}
