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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k, flag = 1;
  cin >> n >> k;
  long long a[n];
  for (long long i = 0; i < n; i++) a[i] = i + 1;
  if (n == k)
    cout << "-1";
  else {
    long long diff = n - 1 - k, i, d;
    if (diff % 2 == 1) {
      swap(a[0], a[1]);
      d = --diff;
      i = 2;
    } else {
      d = diff;
      i = 1;
    }
    while (d != 0) {
      if (i >= n) {
        flag = 0;
        break;
      }
      swap(a[i], a[i + 1]);
      i += 2;
      d -= 2;
    }
    if (flag == 0)
      cout << "-1" << endl;
    else {
      for (long long i = 0; i < n; i++) cout << a[i] << " ";
    }
  }
  return 0;
}
