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<bits/stdc++.h>
using namespace std;

int const maxn = 1e5 + 10;

int N, a[maxn], b[maxn];

int main()
{
  scanf("%d", &N);

  int deg[maxn] = {};
  for(int i = 1; i <= N; i++) {
    scanf("%d", &a[i]);
    deg[a[i]]++;
  }

  bool vis[maxn] = {};
  function<void(int, int)> dfs = [&](int cur, int mn) {
    if(cur == N+1) {
      for(int i = 1; i <= N; i++) printf("%d ", b[i]);
      putchar('\n');
      exit(0);
    }

    if(cur > 1 && !vis[a[b[cur-1]]] && deg[a[b[cur-1]]] == N-cur) return;

    while(vis[mn]) mn++;

    for(int v = mn; v <= N; v++) {
      if(vis[v] || (cur > 1 && v == a[b[cur-1]])) continue;
      b[cur] = v;
      vis[v] = true;
      deg[a[v]]--;
      dfs(cur+1, v==mn? mn+1 : mn);
      deg[a[v]]++;
      vis[v] = false;
    }
  };

  dfs(1, 1);
  puts("-1");

  return 0;
}
