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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k, tempk;
  cin >> n >> k;
  tempk = k;
  if (k >= n)
    cout << -1;
  else if (k == (n - 1)) {
    for (int i = 1; i <= n; i++) cout << i << " ";
  } else {
    int arr[n + 1];
    int even = 2, odd = 3, l = 2;
    while (tempk--) {
      if (l % 2 == 0) {
        arr[l] = even;
        even += 2;
      } else {
        arr[l] = odd;
        odd += 2;
      }
      l++;
    }
    while (l < n) {
      arr[l] = l + 1;
      l++;
    }
    arr[1] = k + 2;
    arr[n] = 1;
    for (int i = 1; i <= n; i++) cout << arr[i] << " ";
  }
  return (0);
}