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
// ※※※ 解答不能 ※※※
// tourist氏.
// https://atcoder.jp/contests/dwacon6th-prelims/submissions/9423322
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define repx(i, a, b) for(int i = a; i < b; i++)
#define rep(i, n) repx(i, 0, n)
#define repr(i, a, b) for(int i = a; i >= b; i--)
const LL MOD = 1e9 + 7;

int main(){
    int n;
    scanf("%d", &n);
    vector<int> a(n), deg(n);
    rep(i, n){
        scanf("%d", &a[i]);
        --a[i];
        deg[a[i]] += 1;
    }
    vector<int> b(n, -1), used(n, 0);
    function<void(int, int)> Dfs = [&](int v, int mn){
        if(v == n){
            rep(i, n){
                if(i > 0) printf(" ");
                printf("%d", b[i] + 1);
            }
            puts("");
            exit(0);
        }
        if(v > 0){
            int who = a[b[v - 1]];
            if(!used[who] && deg[who] == n - v - 1) return 0;
        }
        while(used[mn]) ++mn;
        repx(i, mn, n){
            if(used[i]) continue;
            if(v > 0 && i == a[b[v - 1]]) continue;
            int new_mn = mn;
            if(i == mn) ++new_mn;
            b[v] = i;
            --deg[a[i]];
            used[i] = 1;
            Dfs(v + 1, new_mn);
            used[i] = 0;
            ++deg[a[i]];
            b[v] = -1;
        }
    };
    Dfs(0, 0);
    puts("-1");
    return 0;
}