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 binary_search(long long a[], long long n, long long what) {
  long long left = 0;
  long long right = n - 1;
  long long mid;
  while (left <= right) {
    mid = left + ((right - left) / 2);
    if (a[mid] == what) {
      return mid;
    } else if (a[mid] < what) {
      left = mid + 1;
    } else {
      right = mid - 1;
    }
  }
  return -1;
}
long long lowerBound(long long a[], long long n, long long what) {
  long long left = 0;
  long long right = n - 1;
  long long mid;
  long long ans = -1;
  while (left <= right) {
    mid = left + ((right - left) / 2);
    cout << left << " " << right << " " << mid << endl;
    if (a[mid] >= what) {
      ans = a[mid];
      right = mid - 1;
    } else {
      left = mid + 1;
    }
  }
  return ans;
}
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool isInt(double x) {
  if (ceil(x) == floor(x) && x > 0) {
    return 1;
  } else {
    return 0;
  }
}
void solve() {
  long long n;
  cin >> n;
  ;
  long long k;
  cin >> k;
  ;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    a[i] = i + 1;
  }
  if (k % 2 == 1 && k == n - 1) {
    for (int i = 0; i < n; i++) {
      cout << a[i] << " ";
    }
    cout << endl;
    ;
  } else if (k == n) {
    cout << -1 << "\n";
  } else {
    long long x = n - k;
    long long now = a[0];
    for (long long i = 0; i < x - 1; i++) {
      a[i] = a[i + 1];
    }
    a[x - 1] = now;
    for (int i = 0; i < n; i++) {
      cout << a[i] << " ";
    }
    cout << endl;
    ;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
}