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>
long long z;
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k, z;
  cin >> n >> k;
  if (k >= n) {
    cout << -1;
    return 0;
  }
  long long a[n];
  for (long long i = (long long)0; i < (long long)n; i++) a[i] = i + 1;
  if (n % 2 != k % 2) {
    z = n - k - 1;
    for (long long i = 1; i < n - 1; i += 2) {
      if (z == 0) break;
      swap(a[i], a[i + 1]);
      z -= 2;
    }
  } else {
    z = n - k - 2;
    for (long long i = 1; i < n - 1; i += 2) {
      if (z == 0) break;
      swap(a[i], a[i + 1]);
      z -= 2;
    }
    swap(a[0], a[n - 1]);
  }
  for (long long i = (long long)0; i < (long long)n; i++) cout << a[i] << " ";
}