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 i, n, k, k1, a[100015];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (i = 1; i <= n; ++i) {
    a[i] = i;
  }
  if (n == k) {
    cout << "-1";
  } else {
    k = n - k;
    if (k % 2 == 0) {
      a[1] = n;
      a[n] = 1;
      --k;
    }
    k1 = (k - 1) / 2;
    for (i = 0; i < k1; ++i) {
      swap(a[i * 2 + 2], a[i * 2 + 3]);
    }
    for (i = 1; i <= n; ++i) {
      cout << a[i] << " ";
    }
  }
  return 0;
}