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 debug(x) {cerr << #x << " = " << x << '\n';}
#define PR(a, n) {cerr << #a << " = "; for (int _=1; _<=n; ++_) cerr << a[_] << ' '; cerr << '\n';}
#define PR0(a, n) {cerr << #a << " = "; for (int _=0; _<n; ++_) cerr << a[_] << ' '; cerr << '\n';}
#define FILE_NAME "data"


const int MAXN = 100002;


int n, a[MAXN], degIn[MAXN];


void readInput() {
    cin >> n;

    for (int i = 1; i <= n; ++i)
        cin >> a[i];
}


bool check(vector<int> candidate) {
    for (int i = 1; i < n; ++i) {
        if (candidate[i] == a[candidate[i - 1]])
            return false;
    }

    return true;
}


void solve() {
    if (n == 2) {
        puts("-1");
        return;
    }

    for (int i = 1; i <= n; ++i) {
        degIn[i] = n - 1;
    }

    for (int i = 1; i <= n; ++i) {
        --degIn[a[i]];
    }

    set<int> lexiOrder;
    set<pair<int, int> > degOrder;
    for (int i = 1; i <= n; ++i) {
        lexiOrder.insert(i);
        degOrder.insert({degIn[i], i});
    }

    int offset = 0;
    vector<int> res;
    while (lexiOrder.size() > 3) {
        auto minDegIn = degOrder.begin();
        auto seMinDegIn = degOrder.begin();
        ++seMinDegIn;

        int cur;
        if (minDegIn->first - offset == 0) {
            res.push_back(minDegIn->second);
            cur = res.back();
            degOrder.erase(minDegIn);
            lexiOrder.erase(res.back());
        }
        else {
            if (res.empty() || *lexiOrder.begin() != a[res.back()]) {
                cur = *lexiOrder.begin();
            }
            else {
                auto tmp = lexiOrder.begin();
                ++tmp;
                cur = *tmp;
            }

            res.push_back(cur);
            lexiOrder.erase(cur);
            degOrder.erase({degIn[cur], cur});
        }

        degOrder.erase({degIn[a[cur]], a[cur]});
        ++degIn[a[cur]];
        ++offset;
        degOrder.insert({degIn[a[cur]], a[cur]});
    }

    for (auto x : lexiOrder) {
        res.push_back(x);
    }

    while (!check(res)) {
        next_permutation(res.end() - 3, res.end());
    }

    for (auto x : res)
        cout << x << ' ';
    cout << '\n';
}


int main() {
    #ifdef GLASSES_GIRL
        freopen(FILE_NAME".in", "r", stdin);
    #endif

    readInput();
    solve(); 
}
