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>
const int MAXN = 200000;
int elems[MAXN];
void init(int n) {
  for (int i = 1; i <= n; ++i) elems[i] = i;
}
void solve(int tmp, int n) {
  int cnt = 0;
  int tmpe;
  if (tmp & 1) {
    for (int i = 1; i <= n && cnt < tmp; i += 2) {
      tmpe = elems[i];
      elems[i] = elems[i + 1];
      elems[i + 1] = tmpe;
      if (i == 1)
        cnt++;
      else
        cnt += 2;
    }
  } else {
    for (int i = 2; i <= n && cnt < tmp; i += 2) {
      tmpe = elems[i];
      elems[i] = elems[i + 1];
      elems[i + 1] = tmpe;
      cnt += 2;
    }
  }
}
void output(int n, int k) {
  int tmp = n - 1 - k;
  solve(tmp, n);
  for (int i = 1; i <= n; ++i) {
    if (i == 1)
      printf("%d", elems[i]);
    else
      printf(" %d", elems[i]);
  }
  printf("\n");
}
int main() {
  int n, k;
  while (scanf("%d%d", &n, &k) != EOF) {
    init(n);
    if (k > n - 1)
      printf("-1\n");
    else {
      output(n, k);
    }
  }
  return 0;
}
