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;
int main() {
  long n = 0, k = 0, i = 0, o = -1, e = -1, j = 0, r = 0;
  cin >> n >> k;
  if (k > (n - 1)) {
    cout << "-1";
    return 0;
  }
  long a[n + 1];
  a[1] = 1;
  long odd[n + 1];
  long even[n + 1];
  o = 0;
  odd[o] = 1;
  for (i = 2; i <= n; i++) {
    if (k > 0) {
      a[i] = i;
      k--;
    } else {
      a[i] = -1;
      if (i % 2 == 0) {
        e++;
        even[e] = i;
      } else {
        o++;
        odd[o] = i;
      }
    }
  }
  for (j = o, r = e; j >= 0 && r >= 0; r--, j--) {
    a[odd[j]] = even[r];
    a[even[r]] = odd[j];
    e--;
    o--;
  }
  for (i = 0; i <= o - 1; i++) {
    long temp = odd[i];
    a[odd[i]] = odd[i + 1];
    a[odd[i + 1]] = temp;
  }
  for (i = 0; i <= e - 1; i++) {
    long temp2 = even[i];
    a[even[i]] = even[i + 1];
    a[even[i + 1]] = temp2;
  }
  for (i = 1; i <= n; i++) cout << a[i] << " ";
  return 0;
}
