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;
inline const char nl = '\n';
inline const long long MOD = 1e9 + 7;
inline const long double pi = 3.141592653589793238;
int fscn() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  int sg = 0;
  if (c == '-') {
    c = getchar();
    sg = 1;
  }
  int ret = 0;
  while (c >= '0' && c <= '9') {
    ret = ret * 10 + c - '0';
    c = getchar();
  }
  return sg ? -ret : ret;
}
void solve() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long n = fscn(), k = fscn();
  if (k == n) {
    cout << -1 << nl;
    return;
  }
  long long arr[n + 1];
  for (long long i = 1; i < n + 1; i++) {
    arr[i] = i;
  }
  k = n - k - 1;
  long long i = 2;
  if (k & 1) {
    swap(arr[1], arr[2]);
    i = 3;
    k -= 1;
  }
  while (k) {
    swap(arr[i], arr[i + 1]);
    k -= 2;
    i += 2;
  }
  for (long long i = 1; i < n + 1; i++) {
    cout << arr[i] << " ";
  }
  cout << nl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  solve();
  return 0;
}
