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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const ll oo = 1e17 + 7;
const ll mod = 1e9 + 7, maxn = 1e5 + 100, maxm = 1e5 + 100;
const double PI = acos(-1);
void solve() {
  int n, k;
  cin >> n >> k;
  if (k > n - 1) {
    cout << -1 << endl;
    return;
  }
  vector<int> vs(n);
  for (int i = 0; i < n; i++) vs[i] = i + 1;
  int i = 0, ans = n - 1;
  while (ans > k) swap(vs[i], vs[i + 1]), ans--, i++;
  for (int i = 0; i < n; i++) {
    cout << vs[i] << " \n"[(i == n - 1)];
  }
}
int main() {
  solve();
  return 0;
}
