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<iostream>
#include<iomanip>
#include<cassert>
#include<stdexcept>
#include<utility>
#include<functional>
#include<numeric>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<array>
#include<stack>
#include<queue>
#include<deque>
#include<vector>
#include<complex>
#include<set>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<string>
#include<bitset>
#include<memory>
using namespace std;
using ll=long long;
int n;
vector<int> a;

void dfs(int cnt,set<int>& s,vector<int>& ans,multiset<int>& ms){
    if(cnt==n){
        for(int i=0;i<n;i++) cout<<ans[i]+1<<" ";
        cout<<endl;
        exit(0);
    }
    auto dosome=[&](int v){
        if(!ans.empty() && a[ans.back()]==v) return;
        s.erase(v);
        ms.erase(ms.find(a[v]));
        ans.push_back(v);

        dfs(cnt+1,s,ans,ms);

        ans.pop_back();
        s.insert(v);
        ms.insert(a[v]);
    };
    if(ms.size()>3){
        if(*ms.begin()==*next(ms.rbegin())){
            if(s.count(*ms.begin())){
                dosome(*ms.begin());
                return;
            }
        }
        if(*ms.rbegin()==*next(ms.begin())){
            if(s.count(*ms.rbegin())){
                dosome(*ms.rbegin());
                return;
            }
        }
    }
    int iter=-1;
    while(true){
        auto it=s.upper_bound(iter);
        iter=*it;
        if(it==s.end()) break;
        dosome(*it);
    }
}
int main(){
    cin>>n;
    a.resize(n);
    for(int i=0;i<n;i++) cin>>a[i],a[i]--;

    if(n==2){
        cout<<-1<<endl;
        return 0;
    }
    set<int> s;
    for(int i=0;i<n;i++) s.insert(i);
    multiset<int> ms;
    for(int i=0;i<n;i++) ms.insert(a[i]);
    vector<int> ans;
    dfs(0,s,ans,ms);
    
    return 0;
}
