Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t, z = 0, l, i = 0, len;
  int n, k = 0, m = 0, bam = 0, dan = 0, ans = 0;
  scanf("%d %d", &n, &k);
  int a[n + 5];
  if (n == k) printf("%d", -1), exit(0);
  for (l = n - k + 1; l < n + 1; l++) a[l] = l;
  for (l = 1; l < n - k; l++) a[l] = l + 1;
  a[n - k] = 1;
  for (l = 1; l < n + 1; l++) printf("%d", a[l]), printf("%s", " ");
  return 0;
}