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;
template <class T>
T max(T &a, T &b) {
  return (!(a < b) ? a : b);
}
template <class T>
T min(T &a, T &b) {
  return (a < b ? a : b);
}
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
template <class T>
T mod(T a, T b) {
  return (a < b ? a : a % b);
}
int power(int a, int b, int c) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % c;
    }
    y = (y * y) % c;
    b /= 2;
  }
  return x % c;
}
int main() {
  int n, k;
  cin >> n >> k;
  if (n == k) {
    cout << "-1\n";
    return 0;
  }
  if (n == 1 && k == 0) {
    cout << "1\n";
    return 0;
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++) ans.push_back(i);
  if (k == n - 1) {
    for (typeof(ans.begin()) it = ans.begin(); it != ans.end(); it++)
      cout << *it << " ";
  } else {
    for (int i = k + 1; i < n; i += 2) {
      if (i + 1 < n)
        swap(ans[i], ans[i + 1]);
      else
        swap(ans[0], ans[i]);
    }
    for (typeof(ans.begin()) it = ans.begin(); it != ans.end(); it++)
      cout << *it << " ";
  }
  cout << endl;
  return 0;
}
