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() {
  long int n, k;
  cin >> n >> k;
  if (k > (n - 1))
    cout << -1 << endl;
  else {
    long int res[n];
    for (long int i = 0; i < n; i++) res[i] = i + 1;
    long int temp = n - k;
    if (temp % 2 == 0) {
      long int t = temp / 2;
      for (long int i = 0; i < t; i++) {
        long int one = 2 * i, two = (2 * i) + 1;
        long int temp = res[one];
        res[one] = res[two];
        res[two] = temp;
      }
    } else {
      long int t = temp / 2;
      for (long int i = 0; i < t; i++) {
        long int one = (2 * i) + 1, two = (2 * i) + 2;
        long int temp = res[one];
        res[one] = res[two];
        res[two] = temp;
      }
    }
    for (long int i = 0; i < n; i++) cout << res[i] << " ";
    cout << endl;
  }
  return 0;
}