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 mod = 1000000007;
class Triplet {
 public:
  long long x;
  long long y;
  long long gcd;
};
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
Triplet extendedEuclid(long long a, long long b) {
  if (b == 0) {
    Triplet ans;
    ans.gcd = a;
    ans.x = 1;
    ans.y = 0;
    return ans;
  }
  Triplet smallAns = extendedEuclid(b, a % b);
  Triplet ans;
  ans.gcd = smallAns.gcd;
  ans.x = smallAns.y;
  ans.y = smallAns.x - (a / b) * smallAns.y;
  return ans;
}
long long modulo(long long a, long long m) {
  Triplet ans = extendedEuclid(a, m);
  return ans.x;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k;
  cin >> n >> k;
  if (n == k) {
    cout << "-1";
    return 0;
  }
  if (k == n - 1) {
    for (long long i = 1; i <= n; i++) cout << i << " ";
    return 0;
  }
  cout << n - k << ' ';
  for (long long i = 1; i < n - k; i++) cout << i << " ";
  for (long long i = n - k + 1; i <= n; i++) cout << i << " ";
  return 0;
}