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;
template <typename T>
void SOUT(T what) {
  cout << what << endl;
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  if (k >= n) {
    cout << "-1\n";
    return 0;
  }
  int i = n;
  vector<int> out;
  for (; i >= 1; i--) {
    if (k > 0) {
      out.push_back(i);
      k--;
    } else
      break;
  }
  for (; i >= 2; i -= 2) {
    out.push_back(i - 1);
    out.push_back(i);
  }
  if (i == 1) out.push_back(1);
  reverse((out).begin(), (out).end());
  for (int i = 0; i < (out.size()); ++i) cout << out[i] << " ";
  cout << endl;
}