Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int compare(const void *a, const void *b) { return *(int *)a - *(int *)b; }
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int i;
  if ((n == 1 && k == 1)) {
    printf("-1\n");
    return 0;
  }
  if (n == k) {
    printf("-1\n");
    return 0;
  }
  if ((n - k) % 2 != 0) {
    printf("1 ");
    for (i = 2; i <= ((n - k) - (n - k) % 2); i += 2) {
      printf("%d %d ", i + 1, i);
    }
    for (i = n - k + 1; i <= n; i++) printf("%d ", i);
  } else {
    for (i = 1; i <= (n - k); i += 2) {
      printf("%d %d ", i + 1, i);
    }
    for (i = n - k + 1; i <= n; i++) printf("%d ", i);
  }
  printf("\n");
  return 0;
}