A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).

A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.

You have integer n. Find some lucky permutation p of size n.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.

Output

Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.

Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.

If there are multiple answers, you can print any of them.

Examples

Input

1


Output

1 


Input

2


Output

-1


Input

4


Output

2 4 1 3 


Input

5


Output

2 5 3 1 4 
#include <bits/stdc++.h>
using namespace std;
template <class F, class T>
T convert(F a, int p = -1) {
  stringstream ss;
  if (p >= 0) ss << fixed << setprecision(p);
  ss << a;
  T r;
  ss >> r;
  return r;
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T cube(T x) {
  return x * x * x;
}
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return s == 0 ? 0 : cntbit(s >> 1) + (s & 1);
}
const long double PI = acos(-1.0);
const long double eps = 1e-9;
const int dr[] = {-1, 0, +1, 0};
const int dc[] = {0, +1, 0, -1};
const int inf = (int)1e9 + 5;
const long long linf = (long long)1e16 + 5;
const long long mod = (long long)1e9 + 7;
const int maxn = 10005;
int a[100005];
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  cin >> n;
  if (n % 4 == 2 || n % 4 == 3) {
    cout << -1 << "\n";
    return 0;
  }
  if (n % 4 == 1) a[(n + 1) / 2] = (n + 1) / 2;
  for (int i = 1; i < n / 2; i += 2) {
    a[i] = i + 1;
    a[i + 1] = n - i + 1;
    a[n - i + 1] = n - i;
    a[n - i] = i;
  }
  for (int i = (1); i <= (n); ++i) cout << a[i] << " ";
  return 0;
}
