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;
long long isPrime(long long n);
vector<bool> siev();
long long gcd(long long n1, long long n2);
long long lcm(long long n1, long long n2);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  if (n == k) {
    cout << -1;
    return 0;
  }
  int arr[n + 1];
  for (int i = 1; i <= n; i++) {
    arr[i] = i;
  }
  for (int i = k + 2; i <= n; i += 2) {
    if (i == n) {
      arr[1] = n;
      arr[i] = 1;
    } else {
      arr[i] = i + 1;
      arr[i + 1] = i;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << arr[i] << " ";
  }
}
long long gcd(long long n1, long long n2) {
  long long a = max(n1, n2), b = min(n1, n2);
  while (a % b != 0) {
    long long temp = b;
    b = a % b;
    a = temp;
  }
  return b;
}
long long lcm(long long n1, long long n2) { return (n2 * n1) / gcd(n1, n2); }
vector<bool> siev() {
  vector<bool> siv(100005, 1);
  for (long long i = 2; i * i <= 100005; i++) {
    if (siv[i] == 1) {
      for (long long j = i * i; j <= 100005; j += i) {
        siv[j] = 0;
      }
    }
  }
  siv[0] = 0;
  siv[1] = 0;
  return siv;
}
long long isPrime(long long n) {
  if (n == 1) {
    return 1;
  }
  if (n == 2) {
    return 0;
  }
  int divs = 0;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      divs++;
    }
  }
  if (divs > 0) {
    return 1;
  } else {
    return 0;
  }
}