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[100010];
int main() {
  int n, k;
  while (scanf("%d%d", &n, &k) != EOF) {
    if (n == k) {
      printf("-1\n");
      continue;
    }
    if (k == n - 1) {
      for (int i = 1; i <= n; i++) {
        printf("%d ", i);
      }
      printf("\n");
      continue;
    }
    for (int i = 1; i <= k + 1; i++) {
      a[i] = i;
    }
    a[k + 2] = 1;
    k = k + 3;
    while (k <= n) {
      a[k] = k - 1;
      k++;
    }
    a[1] = n;
    for (int i = 1; i <= n; i++) {
      printf("%d ", a[i]);
    }
    printf("\n");
  }
}