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 gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int n, k;
  bool exit = 1;
  vector<int> sol;
  cin >> n >> k;
  if (k == n) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= k + 1; i++) {
    sol.push_back(i);
  }
  if (k < n - 1) {
    int z;
    if (((k + 2) & 1))
      z = n & 1;
    else
      z = !(n & 1);
    for (int i = k + 2; i <= n - z; i++) {
      if (exit)
        sol.push_back(i + 1);
      else
        sol.push_back(i - 1);
      exit = !exit;
    }
    if (z) {
      sol.push_back(n);
      swap(sol[n - 1], sol[0]);
    }
  }
  for (int i = 0; i < sol.size(); i++) cout << sol[i] << " ";
  return 0;
}