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;
int main() {
  int t, c;
  int a[100009];
  cin >> t >> c;
  if (t == c) {
    printf("-1\n");
  } else {
    if (t - 1 == c) {
      for (int i = 1; i <= t; i++) {
        a[i] = i;
      }
    } else if (c == 0) {
      for (int e = 1; e < t; e++) {
        a[e] = e + 1;
      }
      a[t] = 1;
    } else {
      int i, y = 0;
      for (i = 2;; i++) {
        y++;
        a[i] = i;
        if (y == c) break;
      }
      a[1] = i + 1;
      int r = i + 1;
      for (int j = i + 1; j < t; j++) {
        r++;
        a[j] = r;
      }
      a[t] = 1;
    }
    int y = 0;
    for (int d = 1; d <= t; d++) {
      if (y == 0)
        printf("%d", a[d]);
      else
        printf(" %d", a[d]);
      y++;
    }
    printf("\n");
  }
  return 0;
}
