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>
using namespace std;
int a[100005];
int main() {
  int i, n, k, j;
  while (scanf("%d%d", &n, &k) != EOF) {
    if (n == k) {
      printf("-1\n");
      continue;
    }
    for (i = 1; i <= n; i++) a[i] = i;
    k = n - 1 - k;
    if (k & 1) {
      swap(a[1], a[n]);
      k--;
    }
    for (i = 1, j = 2; i <= k / 2; i++, j += 2) swap(a[j], a[j + 1]);
    printf("%d", a[1]);
    for (i = 2; i <= n; i++) printf(" %d", a[i]);
    printf("\n");
  }
  return 0;
}