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() {
  int n, k;
  cin >> n >> k;
  int i;
  if (n - k < 1) {
    cout << -1 << endl;
    return 0;
  } else if (n - k == 1) {
    for (i = 1; i <= n; i++) {
      cout << i << " ";
    }
    cout << endl;
    return 0;
  }
  if (k == 0) {
    for (i = 1; i < n; i++) {
      cout << i + 1 << " ";
    }
    cout << 1 << endl;
    return 0;
  }
  vector<int> v;
  for (i = 1; i <= k + 1; i++) {
    v.push_back(i);
  }
  int t, l = 0;
  if (n - k == 2) {
    v.push_back(n);
    t = v[0];
    v[0] = v[n - 1];
    v[n - 1] = t;
  } else {
    for (i = k + 2; i <= n; i++) {
      if (l == 0) {
        v.push_back(i + 1);
        l = 1;
      } else {
        v.push_back(i - 1);
        l = 0;
      }
    }
    if ((n - k - 1) % 2 != 0) {
      v[n - 1] = v[n - 1] - 1;
      t = v[0];
      v[0] = v[n - 1];
      v[n - 1] = t;
    }
  }
  for (i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << endl;
}