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 gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int x, y, arr[100010];
  bool ok = false;
  cin >> x >> y;
  if (x == y) {
    cout << -1;
    return 0;
  }
  int index = 1;
  for (int i = 1; i < x; i++) {
    arr[i] = index;
    index++;
  }
  arr[0] = x - y;
  for (int i = 0; i < y; i++) {
    arr[x - i - 1] = x - i;
  }
  for (int i = 0; i < x; i++) cout << arr[i] << " ";
  cout << endl;
  return 0;
}
