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<bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define pll pair<LL,LL>
#define x first
#define y second
#define pi acos(-1)
#define sqr(x) ((x)*(x))
#define pdd pair<double,double>
#define MEMS(x) memset(x,-1,sizeof(x))
#define MEM(x) memset(x,0,sizeof(x))
#define less Less
#define EPS 1e-4
#define arg ARG
#define cpdd const pdd
#define rank Rank
#define KK 500
int main(){
    int n;
    int a[100005];
    int cnt[100005];
    MEM(cnt);
    scanf("%d",&n);
    if(n==2){
        printf("-1\n");
        return 0;
    }
    for(int i = 1;i<=n;i++){
        scanf("%d",&a[i]);
        cnt[a[i]]++;
    }
    priority_queue<pii> pq;
    for(int i = 1;i<=n;i++){
        pq.push(mp(cnt[i],i));
    }
    set<int> s;
    for(int i = 1;i<=n;i++){
        s.insert(i);
    }
    int ans[100005];
  //  printf("?\n");
    for(int i = 1;i<=n;i++){
        ans[i]=0;
      //  printf("%d\n",i);
        while(!pq.empty()&&pq.top().x==n-i){
            if(s.find(pq.top().y)==s.end()){
                pq.pop();
            }
            else if(cnt[pq.top().y]==n-i){
            //    printf("%d %d %d\n",ans[i],pq.top().y,cnt[pq.top().y]);
                assert(ans[i]==0);
                ans[i]=pq.top().y;
                cnt[a[ans[i]]]--;
                pq.push(mp(cnt[a[ans[i]]],a[ans[i]]));
                s.erase(ans[i]);
                pq.pop();
                //if(!pq.empty()&&pq.top().y)
            }
            else{
                pq.pop();
            }
        }
        if(!ans[i]){
            int x=*s.begin();
            s.erase(s.begin());
            if(a[ans[i-1]]!=x){
                ans[i]=x;
                cnt[a[x]]--;
                pq.push(mp(cnt[a[x]],a[x]));
            }
            else{
                ans[i]=*s.begin();
                s.erase(s.begin());
                s.insert(x);
                cnt[a[ans[i]]]--;
                pq.push(mp(cnt[a[ans[i]]],a[ans[i]]));
            }
        }
    }
    for(int i= 1;i<=n;i++)
        printf("%d ",ans[i]);
    printf("\n");
}