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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <map>

#define rep(i,n) for(int i = 0; i < (n); ++i)

#define debug(vec) for(auto v : vec) cout << v << " "; cout << endl;

#define debug2D(vec2D) for(auto vec : vec2D) { for (auto v : vec) cout << v << " "; cout << endl; } 

using namespace std;

typedef long long ll;

const long long int INF = 1 << 30; //10^10
//const ll MOD = 998244353;
const ll MOD = 1000000007;

int gcd(int a,int b) {
  if (a % b == 0) return b;
	else return gcd(b,a % b);
}

int main() {
	ll n;
	cin >> n;

	vector<int> a(n+1);
	rep(i,n) {cin >> a[i+1]; }

	priority_queue<int,vector<int>,greater<int>> que;
	vector<int> b(n+1);

	rep(i,n+1) que.push(i+1);

	for(int i = 1; i <= n; i++) {
	  int k = que.top(); que.pop();
		if (i > 1 && a[b[i-1]] == k) {
			b[i] = que.top();
			que.pop();
			que.push(k);
		}
		else { 
		  b[i] = k;
		}
		if (b[i] >= n+1) { cout << -1 << endl; return 0;}
	}

	rep(i,n) cout << b[i+1] << " "; cout << endl;
}