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 double EPS = 1e-9;
const int MOD = 1000003;
const int INF = 2000 * 1000 * 1000;
using namespace std;
int popcount(int n) {
  bitset<32> b(n);
  return b.count();
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) {
  int temp = gcd(a, b);
  return temp ? (a / temp * b) : 0;
}
template <typename T>
T modpow(T base, T exp, T modulus) {
  base %= modulus;
  T result = 1;
  while (exp > 0) {
    if (exp & 1) result = (result * base) % modulus;
    base = (base * base) % modulus;
    exp >>= 1;
  }
  return result;
}
int main() {
  int n, k;
  cin >> n >> k;
  bool f = 0;
  if (n == k) {
    cout << -1 << endl;
    return 0;
  }
  if (n - k == 1)
    cout << "1 ", f = 1;
  else
    cout << k + 2 << " ";
  for (int i = 2; i <= k + 1; i++) cout << i << " ";
  if (!f) {
    for (int i = k + 3; i <= n; i++) cout << i << " ";
    cout << 1 << endl;
  }
  return 0;
}