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;
void swap(long long int ans[], long long int i, long long int j) {
  long long int temp = ans[i];
  ans[i] = ans[j];
  ans[j] = temp;
}
int main(int argc, char const *argv[]) {
  long long int n, k;
  cin >> n >> k;
  long long int ans[100022];
  for (long long int i = 0; i < 100022; i++) ans[i] = i;
  if (n == k) {
    cout << "-1\n";
    return 0;
  } else {
    if (n % 2 == 0) {
      if (k % 2 == 0) {
        for (long long int i = 1; i <= (n - k); i += 2) swap(ans, i, i + 1);
      } else {
        for (long long int i = 2; i <= (n - k); i += 2) swap(ans, i, i + 1);
      }
    } else {
      if (k % 2 == 0) {
        for (long long int i = 2; i <= (n - k); i += 2) swap(ans, i, i + 1);
      } else {
        for (long long int i = 1; i <= (n - k); i += 2) swap(ans, i, i + 1);
      }
    }
  }
  for (long long int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}