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.
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x7f7f7f7f;
const double eps = 1e-8;
const double PIE = acos(-1.0);
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
const int fx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
const int fy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
void openfile() {
  freopen("data.in", "rb", stdin);
  freopen("data.out", "wb", stdout);
}
int main() {
  int n, k;
  scanf("%d", &n);
  scanf("%d", &k);
  if (n <= k)
    printf("%d", -1);
  else {
    if ((n - k) & 1) {
      printf("%d", 1);
      for (int i = 0; i < k; i++) {
        printf(" ");
        printf("%d", i + 2);
      }
      int cnt = 0;
      for (int i = k + 2; i < n + 1; i++) {
        printf(" ");
        if (cnt & 1)
          printf("%d", i - 1);
        else
          printf("%d", i + 1);
        cnt++;
      }
    } else {
      printf("%d", n);
      for (int i = 0; i < k; i++) {
        printf(" ");
        printf("%d", i + 2);
      }
      int cnt = 0;
      for (int i = k + 2; i < n; i++) {
        printf(" ");
        if (cnt & 1)
          printf("%d", i - 1);
        else
          printf("%d", i + 1);
        cnt++;
      }
      printf(" ");
      printf("%d", 1);
    }
  }
  printf("\n");
  return 0;
}
