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.
#include <bits/stdc++.h>
using namespace std;
long long ans[100005];
int main() {
  long long n, k, i;
  scanf("%lld%lld", &n, &k);
  if (n == k) {
    printf("-1\n");
    return 0;
  }
  for (i = 2; i <= (k + 2 - 1); i++) {
    ans[i] = i;
  }
  long long d = n - i + 1;
  long long lim;
  if (d % 2 == 0)
    lim = n, ans[1] = 1;
  else
    lim = n - 1, ans[1] = n, ans[n] = 1;
  bool f = 1;
  for (; i <= lim; i++) {
    if (f)
      ans[i] = i + 1;
    else
      ans[i] = i - 1;
    f = !f;
  }
  for (i = 1; i <= n; i++) {
    if (i != 1) printf(" ");
    printf("%lld", ans[i]);
  }
  printf("\n");
  return 0;
}
