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>
#define rep(i,l,r) for (int i=l,i##end=r;i<=i##end;++i)
#define sit set<int>::iterator
using namespace std;
const int N=1e6+6;
int a[N],d[N],ans[N];
bool vis[N];
set<int> s;
int main(){
    int n; cin>>n;
    rep(i,1,n) scanf("%d",&a[i]),++d[a[i]];
    rep(i,1,n) s.insert(i);
    rep(i,1,n-3){
        for (sit it=s.begin();it!=s.end();++it)
            if (*it!=a[ans[i-1]]&&(vis[a[*it]]||d[a[*it]]!=n-i)){
                ans[i]=*it; break;
            }
        if (!ans[i]){ cout<<-1<<endl; return 0; }
        vis[ans[i]]=true,--d[a[ans[i]]],s.erase(ans[i]);
    }
    int t=0,b[10];
    rep(i,1,n) if (!vis[i]) b[++t]=i;
    do{
        rep(i,1,t) ans[max(n-3,0)+i]=b[i];
        bool f=false;
        rep(i,max(n-3,0)+1,n)
            if (ans[i]==a[ans[i-1]]){ f=true; break; }
        if (!f){
            rep(i,1,n) printf("%d ",ans[i]);
            return 0;
        }
    }while (next_permutation(b+1,b+1+t));
    cout<<-1<<endl;
    return 0;
}
/*
4
1
*/
