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;

const int N=100000;

int n,a[N+9],deg[N+9];

void into(){
  scanf("%d",&n);
  for (int i=1;i<=n;++i){
	scanf("%d",&a[i]);
	++deg[a[i]];
  }
}


int tmp[N+9],ct;
int vis[N+9];
int ans[N+9];

bool Dfs_ans(int k,int last){
  if (k>n) return 1;
  for (int i=1;i<=ct;++i)
	if (!vis[i]&&a[last]^tmp[i]){
	  vis[i]=1;
	  ans[k]=tmp[i];
	  if (Dfs_ans(k+1,tmp[i])) return 1;
	  vis[i]=0;
	}
  return 0;
}

struct state{
  int x,v;
  state(int X=0,int V=0){x=X;v=V;}
  bool operator < (const state &p)const{return v>p.v||v==p.v&&x<p.x;}
};
set<state>sdeg;
set<int>sid;

void Get_ans(){
  if (n==2) return;
  for (int i=1;i<=n;++i){
	sdeg.insert(state(i,deg[i]));
	sid.insert(i);
  }
  for (int i=1;i<=n-5;++i){
	int t=a[ans[i-1]]==*sid.begin()?*++sid.begin():*sid.begin();
	if (sdeg.begin()->v==n-i) t=sdeg.begin()->x;
	ans[i]=t;
	sdeg.erase(state(t,deg[t]));
	sid.erase(t);
	if (sid.find(a[t])!=sid.end()){
	  sdeg.erase(state(a[t],deg[a[t]]));
	  sdeg.insert(state(a[t],--deg[a[t]]));
	}
  }
  for (set<int>::iterator it=sid.begin();it!=sid.end();++it) tmp[++ct]=*it;
  Dfs_ans(max(n-4,1),ans[max(n-5,0)]);
  ans[0]=1;
}

void work(){
  Get_ans();
}

void outo(){
  if (!ans[0]) puts("-1");
  else{
	for (int i=1;i<=n;++i)
	  printf("%d ",ans[i]);
    puts("");
  }
}

int main(){
  into();
  work();
  outo();
  return 0;
}