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>
//#include <iostream>
//#include <vector>
//#include <string>
//#include <iomanip>
//#include <ctime>


using namespace std;

#define pb push_back
#define mp make_pair
#define eb emplace_back


typedef long long ll;
typedef long double ld;


const ll inf = 2e9;
const ll base = 998244353;
ll mod = 1e9 + 7;


int main() {
//    srand(time(NULL));
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
//    std::random_device rd;
//    std::mt19937 g(rd());
    auto start = clock();

    int n; cin >> n;
    vector<int> key_to_value(n);
    vector<int> cnt(n);
    for (int i = 0; i < n; ++i) {
        cin >> key_to_value[i];
        key_to_value[i]--;
        cnt[key_to_value[i]]++;
    }
    if (n == 2 && key_to_value[0] == 1 && key_to_value[1] == 0) {
        cout << -1;
        return 0;
    }
    set<int> not_used;
    vector<char> used(n, 0);
    vector<int> ans;
    for (int i = 0; i < n; ++i) not_used.insert(i);

    int bad = -1;
    for (int i = 0; i < n - 3; ++i) {
        for (auto c : not_used) {
            if (c == bad) continue;
            if (cnt[key_to_value[c]] >= n - i - 1 && !used[key_to_value[c]]) {
                continue;
            }
            used[c] = 1;
            not_used.erase(c);
            ans.pb(c);
            cnt[key_to_value[c]]--;
            bad = key_to_value[c];
            break;
        }
    }
    pair<pair<int, int>, int> best = {{INT_MAX, INT_MAX}, INT_MAX};
    for (auto i : not_used) {
        for (auto j : not_used) {
            for (auto k : not_used) {
                if ((set<int>{i, j, k}).size() != 3) continue;
                if (bad == i) continue;
                if (j == key_to_value[i]) continue;
                if (k == key_to_value[j]) continue;
                best = min(best, {{i, j}, k});
            }
        }
    }
    ans.pb(best.first.first);
    ans.pb(best.first.second);
    ans.pb(best.second);
    for (auto c : ans) {
        cout << c + 1 << ' ';
    }


    std::cerr << double(clock() - start) / CLOCKS_PER_SEC << std::endl;
    return 0;
}