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 int long long


        signed main() {
        int n,t;
        cin>>n;
        int a[n],ans[n-1];priority_queue<int,vector<int>,greater<int>> que;
        for(int i=0;i<n;i++){
            cin>>a[i];
            a[i]--;
            que.push(i);
        }
        
        for(int i=0;i<n-1;i++){
            t=que.top();
            que.pop();
            if(i!=0&&a[ans[i-1]]==t){
                ans[i]=que.top();
                que.pop();
                que.push(t);
            }else ans[i]=t;
        }
        t=que.top();
        if(a[ans[n-2]]!=t){
            for(int i=0;i<n-1;i++)cout<<ans[i]+1<<' ';
            cout<<t+1<<' ';
            return 0;
        }
        int ok=-1,ng=n-2,mid;
       while(ng>=0&&a[ans[ng]]==t)ng--;
       ng++;
       ok=ng-1;
       
        if(ng!=n-2)sort(ans+ng,ans+ng+2);
        if(a[t]==ans[ng]){
            if(ng!=n-2)swap(ans[ng],ans[ng+1]);
            else{
                if(n==2){
                    cout<<-1;
                    return 0;
                }
                swap(ans[ok],t);
                swap(ans[ok],ans[ng]);
                if(ok!=0&&a[ans[ok-1]]==ans[ok])swap(ans[ok],ans[ng]);
            }
        }
        for(int i=0;i<ng;i++)cout<<ans[i]+1<<' ';
        cout<<t+1<<' ';
        for(int i=ng;i<n-1;i++)cout<<ans[i]+1<<' ';



         return 0;

        }
