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 ld  double

#define sz(x)   (int)x.size()
#define all(x)  x.begin(),x.end()

#define pb  emplace_back
#define X   first
#define Y   second

const int   N   = 2e5 + 5;

typedef pair<int,int>   ii;

int a[N];
int d[N];

int main()  {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    int n;  cin >> n;   a[n] = n;
    set<int> S;
    set<ii>  G;

    for(int i = 0 ; i < n ; ++i)    {
        int x;  cin >> x;   --x;
        a[i] = x;
        d[x]++;
    }
    for(int i = 0 ; i < n ; ++i)
        S.insert(i),
        G.insert(ii(d[i],i));

    vector<int> res;

    int last = n;

    while (S.size())    {
        bool found = 0;
        if (S.size() == 3)   {
            vector<int> P;
            for(int x : S)
                P.pb(x);

            do  {
                if (P[0] == a[last])    continue;
                if (P[1] == a[P[0]])    continue;
                if (P[2] == a[P[1]])    continue;

                for(int x : P)
                    res.pb(x);

                found = 1;

                break;
            }   while (next_permutation(all(P)));

            if(!found)  {
                cout << "-1";
                return  0;
            }
            break;
        }

        for(int x : S)  if (x != a[last])   {
            G.erase (ii(d[x],x));

            if (G.empty() || (*--G.end()).X < sz(S) - 1)    {
                found = 1;
                res.pb(x);
                last = x;
                S.erase(x);
                break;
            }
            G.insert(ii(d[x],x));
        }
        if(!found)  {
            cout << -1;
            return  0;
        }
        int u = a[last];

        if (G.find(ii(d[u],u)) != G.end())  {
            G.erase (ii(d[u],u));   --d[u];
            G.insert(ii(d[u],u));
        }
    }
    for(int x : res)
        cout << x + 1 << " ";
}