612_E. Square Root of Permutation_36763
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;
const int inf = (int)1.01e9;
const double eps = 1e-9;
const int maxn = (int)1e6 + 10;
int a[maxn];
int b[maxn];
int mem[maxn];
int used[maxn];
int total = 0;
void make(vector<int> v) {
  total += ((int)(v).size());
  int step = (((int)(v).size()) + 1) / 2;
  for (int i = 0; (i) < (((int)(v).size())); ++i) {
    int j = (i + step) % ((int)(v).size());
    b[v[i]] = v[j];
  }
}
void make(vector<int> v1, vector<int> v2) {
  int n = ((int)(v1).size());
  total += 2 * n;
  v1.push_back(v1[0]);
  for (int i = 0; (i) < (n); ++i) {
    b[v1[i]] = v2[i];
    b[v2[i]] = v1[i + 1];
  }
}
vector<vector<int> > v;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; (i) < (n); ++i) scanf("%d", a + i);
  for (int i = 0; (i) < (n); ++i) --a[i];
  memset(mem, -1, sizeof mem);
  for (int i = 0; (i) < (n); ++i)
    if (!used[i]) {
      int x = i;
      vector<int> vv;
      while (!used[x]) {
        used[x] = 1;
        vv.push_back(x);
        x = a[x];
      }
      v.push_back(vv);
      if (((int)(vv).size()) & 1) {
        make(vv);
      } else {
        if (mem[((int)(vv).size())] == -1) {
          mem[((int)(vv).size())] = ((int)(v).size()) - 1;
        } else {
          make(vv, v[mem[((int)(vv).size())]]);
          mem[((int)(vv).size())] = -1;
        }
      }
    }
  if (total != n)
    printf("-1\n");
  else {
    for (int i = 0; (i) < (n); ++i) printf("%d%c", b[i] + 1, " \n"[i + 1 == n]);
    for (int i = 0; (i) < (n); ++i) assert(a[i] == b[b[i]]);
  }
  return 0;
}

4
2 1 3 4
 4
2 1 4 3
 5
2 3 4 5 1
 10
3 5 1 2 10 8 7 6 4 9
 100
11 9 35 34 51 74 16 67 26 21 14 80 84 79 7 61 28 3 53 43 42 5 56 36 69 30 22 88 1 27 65 91 46 31 59 50 17 96 25 18 64 55 78 2 63 24 95 48 93 13 38 76 89 94 15 90 45 81 52 87 83 73 44 49 23 82 85 75 86 33 47 19 58 97 37 20 40 10 92 4 6 68 77 54 71 12 62 60 100 39 41 99 72 29 57 8 70 32 66 98
 1
1
 100
94 22 24 99 58 97 20 29 67 30 38 64 77 50 15 44 92 88 39 42 25 70 2 76 84 6 37 49 17 71 31 19 26 79 10 35 65 63 32 95 5 8 52 27 83 18 53 93 13 81 48 68 54 82 34 60 87 23 16 86 55 40 61 45 28 7 74 41 14 91 3 72 33 11 98 89 90 69 78 36 80 59 56 21 43 1 75 46 47 12 96 73 57 51 4 85 9 100 66 62
 10
8 2 10 3 4 6 1 7 9 5
 4
1 2 4 3
 4
1 4 2 3
 4
3 1 2 4
 10
8 3 10 2 4 6 1 7 9 5
 10
3 5 2 1 10 8 7 6 4 9
 5
1 3 4 5 2
 5
4 3 2 5 1

-1
 3 4 2 1  4 5 1 2 3
 6 9 8 10 4 3 7 1 5 2  -1
 1
 78 52 95 76 96 49 53 59 77 100 64 11 9 48 15 17 44 46 32 54 84 68 43 4 21 28 73 6 16 62 31 39 65 86 98 75 33 45 19 3 91 82 2 92 63 88 7 50 97 93 14 22 20 42 60 55 80 85 29 34 56 71 83 38 26 47 90 70 51 41 40 72 37 12 35 99 67 94 1 87 57 8 61 25 23 79 36 18 66 74 5 27 81 69 24 58 13 10 89 30
 -1
 -1
 1 3 4 2
 2 3 1 4
 7 5 4 10 3 6 8 1 9 2
 -1
 -1
 -1
