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;
long long a[100005];
void solve() {
  long long n, k;
  cin >> n >> k;
  if (n == k) {
    cout << -1 << endl;
    return;
  };
  if (k == 0) {
    cout << n << " ";
    for (__typeof((n)) i = (1); i < (n); i++) cout << i << " ";
    return;
  }
  if (k == n - 1) {
    for (__typeof((n + 1)) i = (1); i < (n + 1); i++) cout << i << " ";
    return;
  }
  k = n - k;
  if (k % 2) cout << k << " ";
  for (__typeof((k / 2)) i = (0); i < (k / 2); i++)
    cout << 2 * i + 2 - k % 2 << " " << 2 * i + 1 + k % 2 << " ";
  for (__typeof((n + 1)) i = (k + 1); i < (n + 1); i++) cout << i << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}