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>
#pragma GCC optimise("Ofast")
using namespace std;
const int mod = 1e+9 + 7;
long long binomialCoeff(long long n, long long k) {
  long long res = 1;
  if (n < k) return 0;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
void solve() {
  int n, k, a[100005], t;
  cin >> n >> k;
  if (k == n) {
    cout << -1;
    return;
  }
  for (long long i = 0; i < n; i++) {
    a[i] = i + 1;
  }
  for (int i = 2 + k - 1; i < n; i = i + 2) {
    t = a[i];
    a[i] = a[(i + 1) % n];
    a[(i + 1) % n] = t;
  }
  for (long long i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) solve();
}