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>
#define R register
#define mp make_pair
#define ll long long
#define pii pair<int, int>
using namespace std;
const int N = 110000;

int n, a[N], dgr[N], b[5];
set<int> abl;

template <class T>
inline void read(T &x) {
    x = 0;
    char ch = getchar(), w = 0;
    while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
    while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
    x = w ? -x : x;
    return;
}

int main() {
    read(n);
    if (n == 2) return printf("-1\n"), 0;
    for (R int i = 1; i <= n; ++i) read(a[i]), ++dgr[a[i]], abl.insert(i);
    int lst = 0;
    for (R int i = 1; i <= n - 3; ++i) {
        auto k = abl.begin();
        if (a[lst] == *k) ++k;
        if (abl.find(a[*k]) != abl.end() && dgr[a[*k]] == n - i) k = abl.find(a[*k]);
        printf("%d ", lst = *k), abl.erase(k), --dgr[a[lst]];
    }
    int num = 0;
    while (abl.size()) b[++num] = *abl.begin(), abl.erase(abl.begin());
    do {
        if (b[1] != a[lst] && b[2] != a[b[1]] && b[3] != a[b[2]]) {
            printf("%d %d %d\n", b[1], b[2], b[3]);
            return 0;
        }
    } while (next_permutation(b + 1, b + 4));
    return 0;
}