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>
const long long INF = 2e18;
const double PI = acos(-1.0);
using namespace std;
long long gcd(long long a, long long b) { return (a) ? gcd(b % a, a) : b; }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void WEZaa() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
int main() {
  WEZaa();
  long long m, n, x, y, c, a, b, d;
  cin >> n >> x;
  if (n > x + 1) {
    cout << x + 2 << " ";
    for (int i = 2; i < 2 + x; i++) cout << i << " ";
    for (int i = x + 2; i < n; i++) cout << i + 1 << " ";
    cout << 1 << "\n";
  } else if (n == x + 1)
    for (int i = 0; i < n; i++) cout << i + 1 << " ";
  else
    cout << -1 << "\n";
}