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;
typedef long long ll;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;
template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

const long double EPS = 1e-10;
const long long INF = 1e18;
const long double PI = acos(-1.0L);
//const ll mod = 1000000007;
ll N;
ll a[105000];

int ans[105000];
set<int> inv[105000];
int cnt[105000];
set<int> Rest;

void f(int val) {
    //cerr << "f: " << val << endl;
    int tmp = a[val];
    if(Rest.find(tmp) != Rest.end()) {
        auto itr = inv[cnt[tmp]].lower_bound(tmp);
        inv[cnt[tmp]].erase(itr);
        cnt[tmp]--;
        inv[cnt[tmp]].insert(tmp);
    }
    auto itr = Rest.lower_bound(val);
    Rest.erase(itr);
    itr = inv[cnt[val]].lower_bound(val);
    inv[cnt[val]].erase(itr);
}

bool ok() {
    for(int i = 2; i <= N; i++) {
        if(a[ans[i-1]] == ans[i]) return false;
    }
    return true;
}

int main() {
    //cout.precision(10);
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> N;
    if(N == 2) {
        cout << -1 << endl;
        return 0;
    }
    for(int i = 1; i <= N; i++) Rest.insert(i);
    for(int i = 1; i <= N; i++) {
        cin >> a[i];
        cnt[a[i]]++;
    }
    for(int i = 1; i <= N; i++) {
        inv[cnt[i]].insert(i);
        //cerr << i << " " << cnt[i] << endl;
    }
    for(int i = 1; i <= N - 3; i++) {
        //cerr << i << " " << Rest.size() << endl;
        if(!inv[N-i].empty()) {
            int val = *inv[N-i].begin();
            //cerr << i << " " << val << endl;
            f(val);
            ans[i] = val;
        } else {
            auto itr = Rest.begin();
            int val = *Rest.begin();
            if(i >= 2 and i < N) {
                if(a[ans[i-1]] == val) {
                    itr++;
                    val = *itr;
                }
            }
            //cerr << val << endl;
            f(val);
            ans[i] = val;
        }
    }
    for(int i = N - 2; i <= N; i++) {
        auto itr = Rest.begin();
        ans[i] = *itr;
        Rest.erase(itr);
    }
    do {
        //for(int i = 1; i <= N; i++) cerr << ans[i] << " ";
        //cerr << endl;
        if(ok()) break;
    } while(next_permutation(ans + N - 2, ans + N + 1));
    for(int i = 1; i <= N; i++) {
        cout << ans[i] << " ";
    }
    cout << endl;
    return 0;
}