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
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")

#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <set>
#include <functional>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <bitset>

using namespace std;
using ll = long long;


int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    vector<int> a(n + 1);
    set<int> unused;
    map<int, int> cnt;
    set<pair<int, int>> xyi;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        cnt[a[i]]++;
        unused.insert(i);
    }

    for (auto p : cnt) {
        xyi.insert({ p.second, p.first });
    }

    vector<int> ans(1);
    for (int i = 1; i <= n; i++) {
        for (int x : unused) {
            xyi.erase({ cnt[a[x]], a[x] });
            cnt[a[x]]--;
            xyi.insert({ cnt[a[x]], a[x] });
            if (a[ans.back()] == x) {
                xyi.erase({ cnt[a[x]], a[x] });
                cnt[a[x]]++;
                xyi.insert({ cnt[a[x]], a[x] });
                continue;
            }
            if (xyi.rbegin()->first == n - i - 1 && unused.count(xyi.rbegin()->second) && a[x] == xyi.rbegin()->second) {
                xyi.erase({ cnt[a[x]], a[x] });
                cnt[a[x]]++;
                xyi.insert({ cnt[a[x]], a[x] });
                continue;
            }
            if (n - i == 2 && xyi.rbegin()->first == 1 && unused.count(xyi.rbegin()->second) && unused.count((++xyi.rbegin())->second) &&
               xyi.rbegin()->second != x && (++xyi.rbegin())->second != x) {
                xyi.erase({ cnt[a[x]], a[x] });
                cnt[a[x]]++;
                xyi.insert({ cnt[a[x]], a[x] });
                continue;
            }
            unused.erase(x);
            ans.push_back(x);
            break;
        }
        if (ans.size() == i) {
            cout << "-1\n";
            return 0;
        }
    }

    for (int i = 1; i <= n; i++) {
        cout << ans[i] << " \n"[i == n];
    }
    return 0;
}