A permutation of length n is an array containing each integer from 1 to n exactly once. For example, q = [4, 5, 1, 2, 3] is a permutation. For the permutation q the square of permutation is the permutation p that p[i] = q[q[i]] for each i = 1... n. For example, the square of q = [4, 5, 1, 2, 3] is p = q2 = [2, 3, 4, 5, 1].

This problem is about the inverse operation: given the permutation p you task is to find such permutation q that q2 = p. If there are several such q find any of them.

Input

The first line contains integer n (1 ≤ n ≤ 106) — the number of elements in permutation p.

The second line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the elements of permutation p.

Output

If there is no permutation q such that q2 = p print the number "-1".

If the answer exists print it. The only line should contain n different integers qi (1 ≤ qi ≤ n) — the elements of the permutation q. If there are several solutions print any of them.

Examples

Input

4
2 1 4 3


Output

3 4 2 1


Input

4
2 1 3 4


Output

-1


Input

5
2 3 4 5 1


Output

4 5 1 2 3
#include <bits/stdc++.h>
int next[1000010];
int ans[1000010], check[1000010];
std::vector<int> V, save[1000010];
int main() {
  int a;
  scanf("%d", &a);
  for (int i = 1; i <= a; i++) scanf("%d", &next[i]);
  for (int i = 1; i <= a; i++) {
    if (check[i] == 0) {
      V.clear();
      int p = i;
      while (check[p] == 0) {
        check[p] = 1;
        V.push_back(p);
        p = next[p];
      }
      if (V.size() % 2 == 1) {
        int p = next[V[(V.size() - 1) / 2]];
        for (int j = 0; j < V.size(); j++) {
          ans[V[j]] = p;
          p = next[p];
        }
      } else
        save[V.size()].push_back(i);
    }
  }
  for (int i = 1; i <= a; i++) check[i] = 0;
  for (int i = 2; i <= a; i += 2) {
    if (save[i].size() % 2 == 1) {
      printf("-1");
      return 0;
    }
    for (int j = 0; j < save[i].size(); j += 2) {
      int s = save[i][j];
      int t = save[i][j + 1];
      while (check[s] == 0) {
        check[s] = 1;
        ans[s] = t;
        ans[t] = next[s];
        s = next[s], t = next[t];
      }
    }
  }
  for (int i = 1; i <= a; i++) printf("%d ", ans[i]);
}
