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 = 100005;
int n, a[N], deg[N], pre[N], nxt[N];
bool del[N];
vector<int> p;

void dfs(int u) {
	if (p.size() == n) {
		for (int x : p) printf("%d ", x);
		puts("");
		exit(0);
	}
	if (!del[a[u]] && deg[a[u]] == n - p.size()) return;
	for (int x = nxt[0]; x <= n; x = nxt[x]) {
		if (x == a[u]) continue;
		--deg[a[u]];
		nxt[pre[x]] = nxt[x];
		pre[nxt[x]] = pre[x];
		p.push_back(x);
		del[x] = true;
		dfs(x);
		del[x] = false;
		nxt[pre[x]] = pre[nxt[x]] = x;
		p.pop_back();
		++deg[a[u]];
	}
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
		scanf("%d", a + i), ++deg[a[i]];
	for (int i = 1; i <= n; ++i) pre[i] = i - 1, nxt[i] = i + 1;
	for (int i = 1; i <= n; ++i) {
		del[i] = true, nxt[i - 1] = i + 1, pre[i + 1] = i - 1;
		p.push_back(i), dfs(i), p.pop_back();
		del[i] = false, nxt[i - 1] = pre[i + 1] = i;
	}
	puts("-1");
}
