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 <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <climits>
#include <cmath>
#include <complex>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#include <cstdint>
using namespace std;
typedef long long ll;
#define MP make_pair
#define PB push_back
#define inf 1000000007
#define mod 1000000007
#define rep(i,n) for(int i = 0; i < (int)(n); ++i)

int main(){
    int n;
    cin >> n;
    if(n==2){
        cout << -1 << endl;
        return 0;
    }
    vector<int>a(n+1);
    vector<int>c(n+1);
    vector<int>pre(n+1);
    rep(i,n){
        cin >> a[i+1];
        c[a[i+1]]++;
        pre[a[i+1]] = c[a[i+1]];
    }
    set<pair<int,int>,greater<pair<int,int> > > s;
    set<int> st;
    rep(i,n){
        s.insert(MP(c[i+1],i+1));
        st.insert(i+1);
    }
    vector<int> res;
    for(int i=0;i<n-3;i++){
        auto x = *s.begin();
        int k = x.first;
        int z = x.second;
        if(k==n-i-1){
            res.push_back(z);
            s.erase(x);
            st.erase(z);
            int y = a[z];
            if(st.count(y)){
                s.erase(MP(pre[y],y));
                pre[y]--; 
                s.insert(MP(pre[y],y));
            }
        }else{
            int p = 0;
            if(i!=0)p = a[res[i-1]];
            auto z = st.begin();
            int nx;
            if(p!=(*z)){
                nx = *z;
            }else{
                z++;
                nx = *z;
            }
            res.push_back(nx);
            st.erase(nx);
            s.erase(MP(pre[nx],nx));
            int y = a[nx];
            if(st.count(y)){
                s.erase(MP(pre[y],y));
                pre[y]--; 
                s.insert(MP(pre[y],y));
            }
        }
    }
    // for(auto x:res){
    //     cerr << x << " ";
    // }
    // cerr << endl;
    vector<int> v;
    for(auto x:st){
        v.push_back(x);
    }
    int ng;
    if(n>3) ng = a[res[n-4]];
    else ng = 0;
    do{
        if(v[0]==ng)continue;
        if(v[1]==a[v[0]])continue;
        if(v[2]==a[v[1]])continue;
        res.push_back(v[0]);
        res.push_back(v[1]);
        res.push_back(v[2]);
        break;
    }while(next_permutation(v.begin(),v.end()));
    rep(i,n-1){
        if(a[res[i]]==res[i+1])assert(0);
    }
    for(auto x:res){
    
        cout << x << " ";
    }
    cout << endl;
    return 0;   
}