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>
const long long INF = 1 << 28;
using namespace std;
long long POWER(long long base, long long expo) {
  long long num = 1;
  long long i = expo;
  while (i) {
    num = num * base;
    i--;
  }
  return num;
}
int main() {
  long long t = 1;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    if (k == n)
      cout << -1;
    else if (k == n - 1) {
      for (long long i = 1; i <= n; i++) cout << i << " ";
    } else {
      long long a[n];
      for (long long i = n; i >= n - k + 1; i--) a[i] = i;
      if ((n - k) % 2 == 0) {
        for (long long i = 1; i <= n - k; i += 2) {
          a[i] = i + 1;
          a[i + 1] = i;
        }
      } else {
        for (long long i = 1; i < n - k; i += 2) {
          a[i] = i + 1;
          a[i + 1] = i;
        }
        a[1] = n - k;
        a[n - k] = 2;
      }
      for (long long i = 1; i <= n; i++) cout << a[i] << " ";
    }
  }
  return 0;
}
