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_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  int a[n];
  for (int i = 0; i < n; i++) {
    a[i] = i + 1;
  }
  int t = n - 1;
  if (abs(t - k) % 2 == 1) {
    swap(a[0], a[1]);
    t--;
  }
  int x = 0;
  if (k > t)
    cout << -1 << endl;
  else if (k == t) {
    for (int i = 0; i < n; i++) {
      cout << a[i] << " ";
    }
  } else {
    int end = n - 2;
    while (t) {
      swap(a[end], a[end + 1]);
      end = end - 2;
      t = t - 2;
      if (t == k) {
        x++;
        break;
      }
    }
    if (x == 0)
      cout << -1 << endl;
    else {
      for (int i = 0; i < n; i++) {
        cout << a[i] << " ";
      }
    }
  }
}