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;
using ll = long long;
#ifdef tabr
#include "library/debug.cpp"
#else
#define debug(...) 0
#endif

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    if (n == 2) {
        int a;
        cin >> a;
        cout << -1 << endl;
        return 0;
    }
    vector<int> a(n), b(n, -1);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        a[i]--;
    }
    b[0] = 0;
    priority_queue<int, vector<int>, greater<int>> pq;
    for (int i = 1; i < n; i++) pq.push(i);
    for (int i = 1; i < n - 1; i++) {
        int s = pq.top();
        pq.pop();
        int t = pq.top();
        if (a[b[i - 1]] == s) {
            b[i] = t;
            pq.pop();
            pq.push(s);
        } else {
            b[i] = s;
        }
    }
    int c = pq.top();
    pq.pop();
    if (a[b[n - 2]] != c) {
        b[n - 1] = c;
        for (int i : b) {
            cout << i + 1 << " ";
        }
        cout << endl;
        return 0;
    }
    int pos = n - 2;
    while (pos && a[b[pos - 1]] == c) pos--;
    if (pos == n - 2) {
        vector<int> x = {b[n - 3], b[n - 2], c};
        sort(x.begin(), x.end());
        do {
            if (a[b[n - 4]] == x[0] || a[x[0]] == x[1] || a[x[1]] == x[2]) continue;
            for (int i = 0; i < 3; i++) {
                b[n - 3 + i] = x[i];
            }
            break;
        } while (next_permutation(x.begin(), x.end()));
        for (int i : b) {
            cout << i + 1 << " ";
        }
        cout << endl;
        return 0;
    }
    vector<int> x;
    for (int i = pos; i < n - 1; i++) x.emplace_back(b[i]);
    b[pos] = c;
    sort(x.begin(), x.end());
    for (int i = pos + 1; i < n; i++) b[i] = x[i - pos - 1];
    if (a[c] == b[pos + 1]) swap(b[pos + 1], b[pos + 2]);
    for (int i : b) {
        cout << i + 1 << " ";
    }
    cout << endl;
    return 0;
}