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;
const int N = 100010;
const int M = 1000010;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int PI = cos(-1.0);
const long long OO = 0x3f3f3f3f3f3f3f3fll;
int a[N];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  if (k == n) return printf("-1"), 0;
  k = n - 1 - k;
  for (int i = 1; i <= n; i++) {
    a[i] = i;
  }
  if (k & 1) swap(a[1], a[n]), k--;
  for (int i = 2; k; i += 2) swap(a[i], a[i + 1]), k -= 2;
  for (int i = 1; i <= n; i++) cout << a[i] << " ";
  cout << endl;
  return 0;
}