Problem:
Score : 800 points

Problem Statement

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.

Sample Input 1

4
2 3 4 1

Sample Output 1

1 3 2 4

The arrangement (1,2,3,4) is lexicographically smaller than (1,3,2,4), but is invalid, since it violates the condition "to the immediate right of Card 1 is not Card 2."

Sample Input 2

2
2 1

Sample Output 2

-1

If no arrangements satisfy the conditions, print -1.

Sample Input 3

13
2 3 4 5 6 7 8 9 10 11 12 13 12

Sample Output 3

1 3 2 4 6 5 7 9 8 10 12 11 13

Solution in C++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <functional>
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <string>
#include <vector>
#include <random>
#include <bitset>
#include <queue>
#include <cmath>
#include <stack>
#include <set>
#include <map>
typedef long long ll;
using namespace std;
const ll MOD = 1000000007LL;

int main() {
    cin.sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int n;
    cin >> n;

    vector <int> a(n + 1, 0);
    vector <int> count(n + 1, 0);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        count[a[i]]++;
    }

    int cur = 1;
    for (int i = 1; i <= n; i++) {
        if (count[i] == n) {
            cur = i;
        } else if (count[i] == n - 1 && a[i] != i) {
            cur = i;
        }
    }

    if (n == 2) {
        cout << -1 << '\n';
        return 0;
    } else if (n == 3) {
        vector <int> tmp;
        for (int i = 1; i <= n; i++) tmp.push_back(i);

        do {
            if (a[tmp[0]] != tmp[1] && a[tmp[1]] != tmp[2]) {
                cout << tmp[0] << ' ' << tmp[1] << ' ' << tmp[2] << '\n';
                return 0;
            }
        } while (next_permutation(tmp.begin(), tmp.end()));
    } else if (n == 4) {
        vector <int> tmp;
        for (int i = 1; i <= n; i++) tmp.push_back(i);

        do {
            if (a[tmp[0]] != tmp[1] && a[tmp[1]] != tmp[2] && a[tmp[2]] != tmp[3]) {
                cout << tmp[0] << ' ' << tmp[1] << ' ' << tmp[2] << ' ' << tmp[3] << '\n';
                return 0;
            }
        } while (next_permutation(tmp.begin(), tmp.end()));
    }

    cout << cur << ' ';

    priority_queue <int, vector<int>, greater<int>> q;
    for (int i = 1; i <= n; i++) {
        if (i != cur) q.push(i);
    }

    while (q.size() > 4) {
        int p = q.top();
        q.pop();

        if (a[cur] == p) {
            int p2 = q.top();
            q.pop();
            q.push(p);
            cout << p2 << ' ';
            cur = p2;
        } else {
            cout << p << ' ';
            cur = p;
        }
    }

    vector <int> tmp;
    while (q.size()) {
        tmp.push_back(q.top());
        q.pop();
    }

    do {
        if (a[cur] != tmp[0] && a[tmp[0]] != tmp[1] && a[tmp[1]] != tmp[2] && a[tmp[2]] != tmp[3]) {
            cout << tmp[0] << ' ' << tmp[1] << ' ' << tmp[2] << ' ' << tmp[3] << '\n';
            return 0;
        }
    } while (next_permutation(tmp.begin(), tmp.end()));

    return 0;
}