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 <cstdio>
#include <iostream>
#include <cassert>
#include <string>
#include <algorithm>
#include <cstring>
#include <utility>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <unordered_map>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
const int MAXN = 123456;
int ans[MAXN], indeg[MAXN];
int a[MAXN], rem[5];
int visit[MAXN];
set<pii> pq;
set<int> cand;
int ans_idx = 0, pre = 0;

void solve(int sz) {
    while ((int)cand.size() > sz) {
        int tpcand = *cand.begin(), nxt;
        auto tp = pq.rbegin();
        if (tp->first == (int)cand.size() - 1) {
            int cur = tp->second;
            pq.erase(prev(pq.end()));
            ans[ans_idx++] = cur;
            cand.erase(cur);
            nxt = cur;
        } else if (a[pre] == tpcand) {
            cand.erase(cand.begin());
            int cur = *cand.begin();
            cand.erase(cand.begin());
            ans[ans_idx++] = cur;
            cand.insert(tpcand);
            nxt = cur;
        } else {
            cand.erase(cand.begin());
            ans[ans_idx++] = tpcand;
            nxt = tpcand;
        }
        visit[nxt] = 1;
        if (!visit[a[nxt]]) {
            int to = a[nxt];
            pq.erase(make_pair(indeg[to], to));
            indeg[to]--;
            pq.insert(make_pair(indeg[to], to));
        }
        pq.erase(make_pair(indeg[nxt], nxt));
        pre = nxt;
    }
}

int main() {
    int N;
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) {
        scanf("%d", &a[i]);
        indeg[a[i]]++;
    }
    if (N == 2) return !printf("-1\n");
    for (int i = 1; i <= N; i++) {
        pq.insert(make_pair(indeg[i], i));
        cand.insert(i);
    }
    solve(3);
    for (int i = 0; i < 3; i++) {
        int tp = *cand.begin();
        cand.erase(cand.begin());
        rem[i] = tp;
    }
    int ans1 = MAXN, ans2 = MAXN, ans3 = MAXN;
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 3; j++) {
            if (j == i) continue;
            for (int k = 0; k < 3; k++) {
                if (k == i || k == j) continue;
                if (a[pre] == rem[i]) continue;
                if (a[rem[i]] == rem[j]) continue;
                if (a[rem[j]] == rem[k]) continue;
                if (rem[i] < ans1) {
                    ans1 = rem[i];
                    ans2 = rem[j];
                    ans3 = rem[k];
                } else if (rem[i] == ans1) {
                    if (rem[j] < ans2) {
                        ans1 = rem[i];
                        ans2 = rem[j];
                        ans3 = rem[k];
                    }
                }
            }
        }
    }
    ans[ans_idx++] = ans1;
    ans[ans_idx++] = ans2;
    ans[ans_idx++] = ans3;
    for (int i = 0; i < N; i++)
        printf("%d ", ans[i]);
    puts("");
    return 0;
}
