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 a[110000];
int ha[110000];
int main() {
  int n, k;
  cin >> n >> k;
  if (k >= n) {
    cout << "-1" << endl;
    return 0;
  }
  for (int i = (2); i <= (k + 1); i++) {
    a[i] = i;
    ha[i] = 1;
  }
  for (int i = (k + 2); i <= (n); i++) {
    a[i] = (i + 1) % n;
    ha[(i + 1) % n] = 1;
    if (a[i] == 0) {
      a[i] = n;
      ha[n] = 1;
    }
  }
  for (int i = (1); i <= (n); i++)
    if (!ha[i]) a[1] = i;
  cout << a[1];
  for (int i = (2); i <= (n); i++) cout << " " << a[i];
  cout << endl;
  return 0;
}