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>
using namespace std;
int p[1000001], s[1000001], an[1000001], x[1000001];
vector<int> c, d[1000001];
bool vis[1000001];
int l;
void go(int v) {
  vis[v] = true;
  c.push_back(v);
  if (v != l) go(s[v]);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n + 1; ++i) scanf("%d", x + i);
  for (int i = 1; i < n + 1; ++i) {
    s[x[i]] = x[x[i]];
    p[x[x[i]]] = x[i];
  }
  int sum = n;
  for (int i = 1; i < n + 1; ++i)
    if (!vis[i]) {
      l = p[i];
      go(i);
      int z = c.size();
      if (z & 1) {
        sum -= z;
        int k = (z + 1) >> 1;
        for (int i = 0; i < z; ++i) {
          if (k == z) k = 0;
          an[c[i]] = c[k];
          ++k;
        }
        c.clear();
      } else {
        if (d[z].empty())
          d[z].insert(d[z].begin(), c.begin(), c.end()), c.clear();
        else {
          sum -= z << 1;
          vector<int> &t = d[z];
          int tm = c.back(), j = 0;
          for (int i = 0; i < c.size(); ++i) {
            an[t[j]] = c[i];
            if (++j == c.size()) j = 0;
            an[c[i]] = t[j];
          }
          t.clear();
          c.clear();
        }
      }
    }
  if (sum)
    printf("-1");
  else
    for (int i = 1; i < n + 1; ++i) printf("%d ", an[i]);
}
