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 min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <class T>
T min(T a, T b, T c, T d) {
  return min(a, min(b, min(c, d)));
}
template <class T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
T max(T a, T b, T c, T d) {
  return max(a, max(b, max(c, d)));
}
bool cmp(const int& a, const int& b) { return a > b; }
int f[300007], _f[300007];
int a[300007];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  if (n == k) {
    puts("-1");
    return 0;
  }
  k = n - k;
  if (k % 2) {
    printf("1 ");
    for (int i = (2); i <= (k); i += 2) {
      printf("%d %d ", i + 1, i);
    }
    for (int i = (k + 1); i <= (n); ++i) printf("%d ", i);
  } else {
    for (int i = (1); i <= (k); i += 2) {
      printf("%d %d ", i + 1, i);
    }
    for (int i = (k + 1); i <= (n); ++i) printf("%d ", i);
  }
  return 0;
}
