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;
 
#define ll long long
#define all(aaa) aaa.begin(), aaa.end()

const int N = 1e5 + 5;
int a[N], b[N];

bool solve_slow(int banned, vector<int> &ans, vector<int> v) {
    sort(all(v));

    do {
        if (v[0] == banned)
            continue;

        bool ok = true;
        for (int i = 0; i + 1 < v.size(); i++) {
            if (a[v[i]] == v[i + 1]) {
                ok = false;
                break;
            }
        }
        if (ok) {
            ans.insert(ans.end(), all(v));
            return true;
        }

    } while (next_permutation(all(v)));

    return false;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n;
    cin >> n;

    set<pair<int, int>> st_p;
    set<int> st;

    for (int i = 0; i < n; i++) {
        st.insert(i);
    }

    for (int i = 0; i < n; i++) {
        cin >> a[i];
        a[i]--;
        b[a[i]]++;
    }

    for (int i = 0; i < n; i++) {
        st_p.insert({b[i], i});
    }

    vector<int> ans;
    int banned = -1;

    for (int i = 0; i + 3 < n; i++) {
        if (st_p.rbegin()->first == n - i - 1) {
            ans.push_back(st_p.rbegin()->second);
            st_p.erase(prev(st_p.end()));
            st.erase(ans.back());
        }
        else {
            auto it = st.begin();
            while (banned == *it)
                it++;
            ans.push_back(*it);
            st.erase(it);
        }
        banned = a[ans.back()];
        st_p.erase({b[a[ans.back()]], a[ans.back()]});
        b[a[ans.back()]]--;
        st_p.insert({b[a[ans.back()]], a[ans.back()]});
    }

    vector<int> v;

    for (const int &x : st) {
        v.push_back(x);
    }

    if (solve_slow(banned, ans, v)) {
        for (const int &x : ans) {
            cout << x + 1 << " ";
        }
    }
    else {
        cout << "-1";
    }

    return 0;
}