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>
#pragma comment(linker, "/STACK:1024000000");
using namespace std;
int num[100010];
int main() {
  int n, m, i;
  scanf("%d %d", &n, &m);
  m = n - m;
  if (m == 0) return puts("-1"), 0;
  if (!(m & 1))
    num[1] = n, num[n] = 1;
  else
    num[1] = 1, num[n] = n;
  for (i = 2; i < m; i += 2) num[i] = i + 1, num[i + 1] = i;
  for (; i < n; ++i) num[i] = i;
  for (i = 1; i <= n; ++i)
    printf("%d", num[i]), printf("%c", i == n ? '\n' : ' ');
  return 0;
}