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 <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <set>
#include <map>

using ll = long long;

using namespace std;

int main() {
    int n;
    cin >> n;
    if (n == 2) {
        cout << -1 << '\n';
        return 0;
    }
    int a[n+1]; 
    map<int, set<int>> cnt;
    for (int i=0; i<n; i++) {
        cin >> a[i];
        a[i]--;
        cnt[a[i]].insert(i);
    }
    a[n] = -1;
    for (auto itr = cnt.begin(); itr != cnt.end(); (itr->second.size() ? ++itr : itr = cnt.erase(itr)));
    set<int> s;
    for (int i=0; i<n; i++) s.insert(i);
    int prev = n;
    vector<int> ans;
    for (int i=0; i<n-3; i++) {
        auto citr0 = cnt.begin();
        auto citr1 = ++cnt.begin();
        if (cnt.size() == 2 && citr0->second.size() == 1 && s.count(citr1->first)) {
            prev = citr1->first;
        } else if (cnt.size() == 2 && citr1->second.size() == 1 && s.count(citr0->first)) {
            prev = citr0->first;
        } else {
            auto itr = s.begin();
            if (*itr == a[prev]) itr++;
            prev = *itr;
        }
        ans.push_back(prev);
        s.erase(prev);
        cnt[a[prev]].erase(prev);
        if (cnt[a[prev]].size() == 0) cnt.erase(a[prev]);
    }
    for (auto &si:s) ans.push_back(si);
    do {
        bool success = true;
        int p = prev;
        for (int i=n-3; i<n; i++) {
            if (ans[i] == a[p]) {
                success = false;
                break;
            }
            p = ans[i];
        }
        if (success) break;
    } while (next_permutation(ans.begin() + n-3, ans.end()));
    for (auto &ai:ans) cout << ai + 1 << ' ';
    cout << '\n';
    return 0;
}