Niwango has N cards, numbered 1,2,\ldots,N. He will now arrange these cards in a row.

Niwango wants to know if there is a way to arrange the cards while satisfying all the N conditions below. To help him, determine whether such a way exists. If the answer is yes, also find the lexicographically smallest such arrangement.

* To the immediate right of Card 1 (if any) is NOT Card a_1.
* To the immediate right of Card 2 (if any) is NOT Card a_2.
* \vdots
* To the immediate right of Card N (if any) is NOT Card a_N.

Constraints

* 2 \leq N \leq 10^{5}
* 1 \leq a_i \leq N
* a_i \neq i

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 \ldots a_N


Output

If no arrangements satisfy the conditions, print `-1`. If such arrangements exist, print the lexicographically smallest such arrangement, in the following format:


b_1 b_2 \ldots b_N


Here, b_i represents the i-th card from the left.

Examples

Input

4
2 3 4 1


Output

1 3 2 4


Input

2
2 1


Output

-1


Input

13
2 3 4 5 6 7 8 9 10 11 12 13 12


Output

1 3 2 4 6 5 7 9 8 10 12 11 13
#include <cstdio>
#include <vector>
#include <set>

int main() {
  int n;
  scanf("%d", &n);
  std::vector<int> a(n), deg(n);
  std::set<int> left;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    ++deg[--a[i]];
    left.insert(i);
  }

  int forbid = -1;
  
  auto check = [&](int x) {
    if (x == forbid) return false;
    int y = a[x];
    if (!left.count(y)) {
      if (left.size() == 3) {
        int u = -1, v = -1;
        for (auto &e: left) if (e != x) {
          if (u == -1) u = e;
          else v = e;
        }
        if (a[u] == v && a[v] == u) return false;
      }
      return true;
    }
    if (x == a[y]) {
      if (deg[x] == 1 && deg[y] == left.size() - 1) return false;
    } else {
      int z = a[y];
      if (!left.count(z)) return deg[y] != left.size() - 1;
      if (y != a[z]) return true;
      if (deg[z] == 1 && deg[y] == left.size() - 1) return false;
    }
    return true;
  };

  std::vector<int> b(n, -1);
  for (int i = 0; i < n; ++i) {
    for (auto it = left.begin(); it != left.end(); ++it) {
      if (check(*it)) { b[i] = *it; break; }
    }
    if (b[i] == -1) break;
    forbid = a[b[i]];
    --deg[forbid];
    left.erase(b[i]);
  }
  if (b.back() == -1) puts("-1");
  else {
    for (auto &x: b) printf("%d ", x + 1);
    puts("");
  }
  return 0;
}
