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;
bool sortpairs(const pair<long long int, long long int> a,
               const pair<long long int, long long int> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first < b.first;
}
void solve() {
  long long int a = 1, b = 0, c = 0, t = 0, k = -1, c1 = 0, c2 = 0, c3, c4,
                n = 0, m = 0, x = 2231131199;
  cin >> a >> b;
  if (a == b) {
    cout << -1;
    return;
  }
  std::vector<long long int> v(a);
  for (int i = 1; i < a + 1; ++i) {
    v[i - 1] = i;
  }
  for (int i = b + 1; i < a - 1; i += 2) {
    swap(v[i], v[i + 1]);
  }
  if (b != a - 1) swap(v[0], v[a - 1]);
  for (auto i : v) cout << i << " ";
  ;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  clock_t z = clock();
  long long int qc = 1;
  for (long long int i = 1; i <= qc; i++) solve();
  fprintf(stderr, "Total Time: % .4Lf\n",
          (long double)(clock() - z) / CLOCKS_PER_SEC),
      fflush(stderr);
}
