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>
#pragma GCC optimize("O3")
using namespace std;
const int mod = 1e7 + 7;
istream &in(cin);
ostream &out(cout);
long long minn = INT_MAX;
int main() {
  long long n, m;
  cin >> n >> m;
  vector<long long> v;
  if (n - m == 0) {
    return cout << -1, 0;
  }
  v.push_back(1);
  long long c = 0, q;
  for (int i = 1; i < n; i++, m--) {
    if (m > 0)
      v.push_back(i + 1);
    else {
      c++;
      if (c == 1) {
        q = i + 1;
      }
      v.push_back(i + 2);
    }
  }
  if (v[v.size() - 1] > n) {
    v[v.size() - 1] = 1;
    v[0] = q;
  }
  for (auto i : v) cout << i << " ";
  return 0;
}
