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[100005];
int main() {
  int n, k, p, current = 2;
  cin >> n >> k;
  p = n - 1;
  if (k == n) {
    cout << -1;
    return 0;
  } else if (k == n - 1) {
    for (int i = 1; i <= n; ++i) {
      cout << i << ' ';
    }
    return 0;
  } else if (k == 0) {
    for (int i = 2; i <= n; ++i) {
      cout << i << ' ';
    }
    cout << 1 << ' ';
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    a[i] = i;
  }
  int v = p - k;
  if (v % 2 == 1) {
    swap(a[1], a[2]);
    p--;
  }
  int i = (p == n - 1) ? 2 : 3;
  for (; i <= n; ++i) {
    if (p == k) {
      for (int j = 1; j <= n; ++j) cout << a[j] << ' ';
      return 0;
    } else {
      swap(a[i], a[i + 1]);
      i++;
      p -= 2;
    }
  }
  cout << -1;
}