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 cnt(long long n) {
  long long c = 0;
  while (n) {
    n &= n - 1;
    c++;
  }
  return c;
}
int main() {
  long long t = 1, n, m, k, x1, x2, x3, x, y, y1, y2, y3;
  while (t--) {
    cin >> n >> k;
    if (k == n) {
      cout << -1;
      return 0;
    }
    vector<int> v;
    for (int i = 2; i <= k + 1; i++) {
      v.push_back(i);
    }
    if (k == n - 1) {
      cout << 1 << " ";
      for (auto i : v) cout << i << " ";
    } else {
      v.insert(v.begin(), n);
      v.push_back(1);
      for (int i = k + 2; i < n; i++) v.push_back(i);
      for (auto i : v) cout << i << " ";
    }
    return 0;
  }
}