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;
int shuzi[100009];
int main() {
  int a, b;
  while (cin >> a >> b) {
    int cnt = 0;
    memset(shuzi, 0, sizeof(shuzi));
    for (int i = 2; i <= a; i++) {
      if (cnt == b) break;
      shuzi[i] = i;
      cnt++;
    }
    if (cnt != b)
      cout << "-1" << endl;
    else {
      int xx = 1;
      for (int i = 2; i <= a; i++) {
        if (shuzi[i] == 0 && xx == 1) {
          xx = i;
        }
        if (shuzi[i] == 0) shuzi[i] = i + 1;
      }
      shuzi[1] = xx;
      if (xx != 1) {
        if (shuzi[a] == a + 1) {
          shuzi[a] = 1;
        }
      }
      cout << shuzi[1];
      for (int i = 2; i <= a; i++) {
        cout << " " << shuzi[i];
      }
      cout << endl;
    }
  }
  return 0;
}
