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;
using namespace std;
long long pow_m(long long x, long long y, long long m = 1000000007) {
  x = x % m;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= m;
    y = y >> 1;
    x = x * x;
    x %= m;
  }
  return res;
}
long long modInverse(long long a, long long m) { return pow_m(a, m - 2, m); }
long long bin_coff(long long n, long long k, long long m) {
  vector<vector<long long>> ans(n + 1, vector<long long>(k, 0));
  for (long long i = 0; i < n + 1; i++) {
    for (long long j = 0; j < min(i, k) + 1; j++) {
      if (j == 0 || j == i)
        ans[i][j] = 1;
      else
        ans[i][j] = ans[i - 1][j - 1] % m + ans[i - 1][j] % m;
    }
  }
  return ans[n][k] % m;
}
long long inverse(long long i) {
  if (i == 1) return 1;
  return (1000000007 -
          ((1000000007 / i) * inverse(1000000007 % i)) % 1000000007 +
          1000000007) %
         1000000007;
}
const long long N = 10000007;
vector<bool> primes(N + 1, true);
void eren() {
  primes[0] = false;
  primes[1] = false;
  for (long long i = 2; i * i <= N; i++) {
    if (primes[i]) {
      for (long long j = i * i; j <= N; j += i) primes[j] = false;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  eren();
  int tt = 1;
  for (int it = 0; it < tt; it++) {
    long long n, k;
    cin >> n >> k;
    if (n == k) {
      cout << "-1\n";
      return 0;
    }
    cout << n - k << " ";
    for (long long i = 1; i < n - k; i++) cout << i << " ";
    for (long long i = n - k + 1; i <= n; i++) cout << i << " ";
  }
  return 0;
}