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>
int main(void) {
  unsigned int n, k, i;
  scanf("%u %u\n", &n, &k);
  assert(1 <= n && n <= 100000 && k <= n);
  if (k == n) {
    printf("%i\n", -1);
  } else if (k == n - 1) {
    for (i = 1; i < n; ++i) {
      printf("%u ", i);
    }
    printf("%u\n", n);
  } else {
    printf("%u ", n - k);
    for (i = 2; i < n - k; ++i) {
      printf("%u ", i - 1);
    }
    printf("%u", n - k - 1);
    if (k > 0) {
      for (i = n - k + 1; i < n + 1; ++i) printf(" %u", i);
    }
    printf("\n");
  }
  return 0;
}
