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
/**
 *    author:  otera    
**/
#include<iostream>
#include<string> 
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<algorithm> 
#include<functional>
#include<iomanip>
#include<queue>
#include<deque>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
using namespace std;

//#define int long long
typedef long long ll;
typedef unsigned long long ul;
typedef unsigned int ui;
typedef long double ld;
const int inf=1e9+7;
const ll INF=1LL<<60 ;
const ll mod=1e9+7 ;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef complex<ld> Point;
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<int, int> P;
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;
#define fr first
#define sc second
#define all(c) c.begin(),c.end()
#define pb push_back
#define debug(x)  cerr << #x << " = " << (x) << endl;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

void fail() {
    cout << -1 << endl;
    exit(0);
}

void out(vector<int> x) {
    rep(i, (int)x.size()) {
        if(i) cout << " ";
        cout << x[i] + 1;
    }
    cout << endl;
    exit(0);
}

void solve() {
    int n;
	cin >> n;
    vector<int> a(n);
    rep(i, n) {
        cin >> a[i];
        -- a[i];
    }
    set<int> se;
    for(int i = 1; i < n; ++ i) se.insert(i);
    vector<int> ans(1, 0);
    for(int i = 1; i < n - 1; ++ i) {
        auto ite = se.begin();
        if(*ite == a[ans.back()]) ++ ite;
        ans.push_back(*ite);
        se.erase(ite);
    }
    int b = *se.begin();
    if(b != a[ans.back()]) {
        ans.push_back(b);
        out(ans);
    }
    set<int> pool;
    while(ans.size() and a[ans.back()] == b) {
        pool.insert(ans.back());
        ans.pop_back();
    } 
    if(pool.size() == 1) {
        if(n == 2) fail();
        vector<int> p;
        p.push_back(b);
        p.push_back(*pool.begin());
        p.push_back(ans.back());
        ans.pop_back();
        sort(all(p));
        auto tmp = ans;
        vector<int> res = {n + 1};
        do {
            for(int x: p) tmp.push_back(x);
            bool ok = 1;
            rep(i, n - 1) {
                if(tmp[i + 1] == a[tmp[i]]) ok = 0;
            }
            if(ok) chmin(res, tmp);
            rep(i, 3) tmp.pop_back();
        } while(next_permutation(all(p)));
        ans = res;
    } else {
        ans.push_back(b);
        while(pool.size()) {
            auto ite = pool.begin();
            if(*ite == a[ans.back()]) ++ ite;
            ans.push_back(*ite);
            pool.erase(ite);
        }
    }
    out(ans);
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(10);
	//int t; cin >> t; rep(i, t)solve();
	solve();
    return 0;
}