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
/// https://atcoder.jp/contests/dwacon6th-prelims/tasks/dwacon6th_prelims_d
#include <bits/stdc++.h>
using namespace std;
#define in ({int x=0;int c=getchar(),n=0;for(;!isdigit(c);c=getchar()) n=(c=='-');for(;isdigit(c);c=getchar()) x=x*10+c-'0';n?-x:x;})
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rng(1);
int rnd(int l,int r){return l+rng()%(r-l+1);}
#define fasty ios_base::sync_with_stdio(0),cin.tie(0);
#define forinc(a,b,c) for(int a=b,_c=c;a<=_c;++a)
#define fordec(a,b,c) for(int a=b,_c=c;a>=_c;--a)
#define forv(a,b) for(auto&a:b)
#define fi first
#define se second
#define pb push_back
#define ii pair<int,int>
#define mt make_tuple
#define all(a) a.begin(),a.end()
#define reset(f, x) memset(f, x, sizeof(f))
#define bit(x,i) ((x>>(i-1))&1)
#define on(x,i) (x|(1ll<<(i-1)))
#define off(x,i) (x&~(1<<(i-1)))
#define gg exit(0);

const int N=100010;

int n;
int cnt[N],a[N];
set<int> st;
vector<int> ans;

void inc(int x){
    ans.pb(x);
    st.erase(x);
}
void dec(){
    st.insert(ans.back());
    ans.pop_back();
}

void dfs(int ban){
    if(ans.size()==n){
        forv(i,ans) cout<<i<<" ";
        gg
    }
    if(st.count(ban) && cnt[ban]==n-ans.size())
        return;
    cnt[ban]--;
    for(auto it=st.begin();it!=st.end();it++){
		if(*it==ban) continue;
		inc(*it);
		dfs(a[*it]);
		dec();
	}
	cnt[ban]++;
}

int main(){
    #define task "arrangement"
    //freopen(task".inp","r",stdin);
    //freopen(task".out","w",stdout);

    if((n=in)==2) return cout<<-1,0;
    forinc(i,1,n) cnt[a[i]=in]++,st.insert(i);
    dfs(0);
}

