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;
const long long OO = (long long)1e15;
vector<long long> v1;
map<long long, long long> ma;
set<long long> s1;
int main() {
  int n, m, i, s = 0;
  cin >> n >> m;
  int a[n];
  if (m > n - 1) {
    cout << -1;
    return 0;
  }
  for (i = 0; i < n; i++) a[i] = i + 1;
  m = n - m;
  m--;
  s = m % 2;
  m = m / 2;
  for (i = 2; i < n; i += 2) {
    if (m == 0) {
      break;
    } else {
      swap(a[i], a[i - 1]);
      m--;
    }
  }
  if (s == 1) {
    swap(a[n - 1], a[0]);
  }
  for (i = 0; i < n; i++) cout << a[i] << " ";
  return 0;
}