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
// Generated by 1.1.6 https://github.com/kyuridenamida/atcoder-tools

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

const i64 MOD = 1e9 + 7;
const i64 INF = i64(1e18) + 7;


template <typename T>
bool chmin(T& x, T y){
    if(x > y){
        x = y;
        return true;
    }
    return false;
}

template <typename T>
bool chmax(T& x, T y){
    if(x < y){
        x = y;
        return true;
    }
    return false;
}

int n;
vector<int> a;
vector<int> path;
multiset<int> ign;
set<int> used;
clock_t st;

void solve(int idx, int now){
    path[idx] = now;
    if((double(clock()) - st) / CLOCKS_PER_SEC >= 2){
        cout << -1 << endl;
        exit(0);
    }
    if(idx == n - 1){
        for(int i = 0; i < n; ++i)
            cout << path[i] + 1 << " \n"[i == n - 1];
        exit(0);
    }

    used.erase(now);

    if(ign.size() >= 2){
        int ign_beg = *ign.begin();
        if(used.find(ign_beg) != used.end()){
            ign.erase(ign.find(a[ign_beg]));
            if(*ign.begin() == *ign.rbegin() && used.find(*ign.begin()) != used.end()){
                used.emplace(now);
                ign.insert(a[ign_beg]);
                return;
            }
            ign.insert(a[ign_beg]);
        }

        int ign_rbeg = *ign.rbegin();
        if(used.find(ign_rbeg) != used.end()){
            ign.erase(ign.find(a[ign_rbeg]));
            if(*ign.begin() == *ign.rbegin() && used.find(*ign.begin()) != used.end()){
                used.emplace(now);
                ign.insert(a[ign_rbeg]);
                return;
            }
            ign.insert(a[ign_rbeg]);
        }
    }


    ign.erase(ign.find(a[now]));

    for(auto it = used.begin(); it != used.end();){
        int val = *it;
        if(val != a[now])
            solve(idx + 1, val);
        it = used.upper_bound(val);
    }
    used.emplace(now);
    ign.insert(a[now]);
}


signed main(){
    long long N;
    st = clock();
    scanf("%lld",&N);
    std::vector<int> A(N);
    for(int i = 0 ; i < N ; i++){
        scanf("%d",&A[i]);
    }
    n = move(N);
    a = move(A);
    path.assign(n, 150);
    for(int i = 0; i < n; ++i){
        used.emplace(i);
        --a[i];
        ign.insert(a[i]);
    }
    for(int i = 0; i < n; ++i)
        solve(0, i);
    cout << -1 << endl;
    return 0;
}
