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 double E = 2.7182818284590452353602874713527;
using namespace std;
long ara[1000000], ara1[1000000];
int main() {
  long n, k, i, sum = 0, y = 0, res;
  scanf("%ld%ld", &n, &k);
  if (n == k) {
    printf("-1\n");
    return 0;
  } else {
    res = n - (k + 1);
    if (((res) % 2 == 0)) {
      for (i = k + 2; i < n; i += 2) {
        ara[i] = i + 1;
        ara[i + 1] = i;
      }
    } else {
      for (i = k + 2; i < n - 1; i += 2) {
        ara[i] = i + 1;
        ara[i + 1] = i;
      }
      ara[1] = n;
      ara[n] = 1;
    }
  }
  for (i = 1; i <= n; i++) {
    if (ara[i] == 0) {
      printf("%ld", i);
    } else {
      printf("%ld", ara[i]);
    }
    if (i != n) {
      printf(" ");
    }
  }
  printf("\n");
  return 0;
}
