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;

const int N = 1e5 + 10;
int deg[N], a[N], n;

bool vis[N];
vector<int> c;

void dfs(int nw, int lst) {
    if (nw >= n) {
        for (auto v : c) cout << v << " ";
        cout << '\n';
        exit(0);
    }
    if (c.size()) {
        int u = a[c.back()];
        if (!vis[u] && nw + deg[u] + 1 == n) return;
    }
    while (lst <= n && vis[lst]) ++lst;
    for (int i = lst; i <= n; ++i) if (!vis[i]) {
        if (nw && i == a[c[nw - 1]]) continue;
        vis[i] = 1;
        c.push_back(i);
        --deg[a[i]];
        dfs(nw + 1, lst);
        vis[i] = 0;
        c.pop_back();
        ++deg[a[i]];
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        ++deg[a[i]];
    }
    dfs(0, 1);
    cout << "-1\n";
    return 0;
}