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 dx[] = {-1, 0, 0, 1, -1, -1, 1, 1};
int dy[] = {0, -1, 1, 0, -1, 1, -1, 1};
const double eps = 1e-9;
int table[100001];
using namespace std;
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  if ((n == k))
    printf("-1");
  else {
    for (int i = 2; i <= k + 1; ++i) {
      table[i] = i;
    }
    int cnt = (n - k);
    if (cnt % 2 == 0) {
      table[1] = n;
      table[n] = 1;
      for (int i = k + 2; i <= n - 1; ++i) {
        table[i] = i;
      }
      for (int i = k + 2; i <= n - 1; i += 2) {
        int t = table[i];
        table[i] = table[i + 1];
        table[i + 1] = t;
      }
    } else {
      table[1] = 1;
      for (int i = k + 2; i <= n; ++i) {
        table[i] = i;
      }
      for (int i = k + 2; i <= n; i += 2) {
        int t = table[i];
        table[i] = table[i + 1];
        table[i + 1] = t;
      }
    }
    for (int i = 1; i <= n; ++i) {
      printf("%d ", table[i]);
    }
  }
  printf("\n");
  return 0;
}