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;
const int mod = 1000000007;
const int INF = 1000000000;
int adj[100005];
int vis[100005];
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int n, m;
  cin >> n >> m;
  vector<int> v;
  if (n == m) {
    printf("-1");
  } else if (n == 1 && m == 0) {
    cout << "1\n";
  } else if (n == 2 && m == 1) {
    cout << "1 2\n";
  } else {
    int p = n;
    for (int i = 1; i <= m; i++) {
      v.push_back(p);
      p--;
    }
    int x = p;
    p--;
    for (int i = v.size() + 1; i <= n; i++) {
      if (i == n) {
        v.push_back(x);
      } else {
        v.push_back(p);
        p--;
      }
    }
    reverse(v.begin(), v.end());
    for (auto x : v) {
      cout << x << " ";
    }
  }
  return 0;
}
