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>
using namespace std;
const int MOD = 1e9 + 7;
const int nax = 1e6 + 10;
long long int i, j, x, y, k, z, cnt;
long long int f = 0;
int main() {
  long long int t = 1;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    vector<long long int> a;
    for (i = 1; i <= n; i++) {
      a.push_back(i);
    }
    if (k >= n) {
      cout << -1 << endl;
    } else if (k == n - 1) {
      for (i = 1; i <= n; i++) cout << i << " ";
      cout << endl;
    } else {
      if ((k % 2 == 0 && n % 2 != 0) || (k % 2 != 0 && n % 2 == 0)) {
        for (i = 1; i < n - k; i += 2) {
          swap(a[i], a[i + 1]);
        }
      } else {
        for (i = 0; i < n - k; i += 2) {
          swap(a[i], a[i + 1]);
        }
      }
      for (i = 0; i < n; i++) {
        cout << a[i] << " ";
      }
      cout << endl;
    }
  }
}
