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;

#define N 100005

int n, a[N];

int b[N], cnt[N];
bool vis[N];

int start;

double TL = 2.35;

using pii = pair <int, int>;

set <pii> st;

void dfs(int sz, int mn) {
	if (sz == n+1) {
        for (int i = 1; i < sz; i ++) printf("%d ", b[i]);
        exit(0);
	}
    if (1.0 * (clock() - start) / CLOCKS_PER_SEC > TL) {
        puts("-1");
        exit(0);
    }
    pii cur = *st.rbegin();
    if (cur.first == n - sz) {
        if (sz == 1 || sz > 1 && a[b[sz-1]] != cur.second) {
            int i = cur.second;
			b[sz] = i;
			st.erase(pii(cnt[i], i));
			if (!vis[a[i]]) {
				st.erase(pii(cnt[a[i]], a[i]));
				cnt[a[i]] --;
				st.insert(pii(cnt[a[i]], a[i]));
			}
			vis[i] = 1;
			dfs(sz+1, mn);
			vis[i] = 0;
			if (!vis[a[i]]) {
				st.erase(pii(cnt[a[i]], a[i]));
				cnt[a[i]] ++;
				st.insert(pii(cnt[a[i]], a[i]));
			}
			st.insert(pii(cnt[i], i));
        }
        else return;
    }
    else {
        while (vis[mn]) mn ++;
        for (int i = mn; i <= n; i ++) if (!vis[i]) {
            if (sz == 1 || sz > 1 && a[b[sz-1]] != i) {
                b[sz] = i;
                st.erase(pii(cnt[i], i));
                if (!vis[a[i]]) {
                    st.erase(pii(cnt[a[i]], a[i]));
                    cnt[a[i]] --;
                    st.insert(pii(cnt[a[i]], a[i]));
                }
                vis[i] = 1;
				dfs(sz+1, mn);
				vis[i] = 0;
                if (!vis[a[i]]) {
                    st.erase(pii(cnt[a[i]], a[i]));
                    cnt[a[i]] ++;
                    st.insert(pii(cnt[a[i]], a[i]));
                }
				st.insert(pii(cnt[i], i));
            }
        }
    }
    return;
}

int main() {
	start = clock();
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++) scanf("%d", &a[i]);
    bool flag = 0;
    int p = -1;
    for (int i = 1; i <= n; i ++) cnt[a[i]] ++;
	for (int i = 1; i <= n; i ++) st.insert(pii(cnt[i], i));
	dfs(1, 1);
    puts("-1");
	return 0;
}
