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 long double ld;
typedef vector<int> vi;
typedef pair<int,int> pi;
typedef vector<pi> vpi;
typedef long double ld;
#define pb emplace_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define ALL(x) x.begin(), x.end()
#define SZ(x) (int)x.size()
#define f first
#define s second
#define MAXN 101000

vi out;
set<int> occ[MAXN];
int blk[MAXN];
int N;
set<int> S;
stack<int> stk;
int done [MAXN];
int O[MAXN];

int main(){
	cin>>N;
	for (int i=1;i<=N;++i){
		cin>>blk[i];
		O[blk[i]]++;
		S.insert(i);
	}
	for (int i=1;i<=N;++i)occ[O[i]].insert(i);

	out.pb(0);
	while (N - SZ(out) > 3){
		int L = N - SZ(out);
		int t = -1;
		if (SZ(occ[L])){
			t = *occ[L].begin();
		}else{
			while (1){
				t = *(S.begin());S.erase(t);
				if (blk[out.back()] == t){
					stk.push(t);
					continue;
				}
				else break;
			}
		}
		// cout<<"Insert "<<t<<'\n';
		out.pb(t);
		if (S.find(t) != S.end())S.erase(t);
		int nxt = blk[t];
		occ[O[nxt]].erase(nxt);
		--O[nxt];
		occ[O[nxt]].insert(nxt);
		while(SZ(stk))S.insert(stk.top()), stk.pop();
	}
	vi tmp;tmp.clear();
	vi cur;cur.clear();
	for (auto i:S)cur.pb(i);
	// for (int i=1;i<=N;++i)cout<<i<<' '<<blk[i]<<'\n';
	do{
		// for (auto i :cur)cout<<i<<' ';cout<<'\n';
		int ok = 1;
		for (int i=1;i<SZ(cur);++i){
			// cout<<blk[cur[i-1]]<<' '<<cur[i]<<'\n';
			if (blk[cur[i-1]] == cur[i])ok = 0;
		}
		// cout<<blk[out.back()]<<' '<<cur[0]<<'\n';
		if (blk[out.back()] == cur[0])ok = 0;
		if (!ok)continue;
		for (auto i : cur)out.pb(i);
		break;
	}while (next_permutation(ALL(cur)));

	if (SZ(out) != N+1)cout<<-1;
	else{
		for (auto i : out)if(i)cout<<i<<' ';
	}
}