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[200000], N, K, total;
void input() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; ++i) {
    a[i] = i;
  }
  total = N - 1;
  if (K < 0 || K > N - 1) {
    printf("-1\n");
    exit(0);
  }
  if (K == 0) {
    printf("%d ", N);
    for (int i = 1; i < N; i++) printf("%d ", i);
    exit(0);
  }
  int to_swap = total - K;
  if (to_swap % 2 != 0) {
    swap(a[1], a[2]);
    to_swap--;
  }
  to_swap /= 2;
  for (int i = 1; i <= to_swap; ++i) {
    swap(a[i * 2 + 1], a[i * 2 + 2]);
    if (i * 2 + 2 > N) {
      printf("-1\n");
      exit(0);
    }
  }
  for (int i = 1; i <= N; ++i) {
    printf("%d ", a[i]);
  }
}
int main() {
  input();
  return 0;
}