[CF 612E]Square Root of Permutation

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

题目大意：

给你个置换p，然后做平方运算，得到置换q，题目给你q，问你能否找到p，要构造出来。

题解：

这道题要求倒推出一个置换，由于原置换p中的环不一定全是奇数环，所以平方之后有可能有环会裂开。

对于平方后的置换q中的奇数环，直接在里面推。偶数环就看是否有相同大小的偶数环与它合并。

//Never forget why you start
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
int n,m,a[1000005],lm,ans[1000005],q[1000005];
struct node{
  int sum;
  vector<int>p;
  friend bool operator < (const node a,const node b){
    return a.sum<b.sum;
  }
}s[1000005];
int vis[1000005],cnt;
void dfs(int r){
  vis[r]=1;
  cnt++;
  s[lm].p.push_back(r);
  if(vis[a[r]])return;
  else dfs(a[r]);
}
int main(){
  int i,j;
  scanf("%d",&n);
  for(i=1;i<=n;i++)scanf("%d",&a[i]);
  for(i=1;i<=n;i++)
    if(!vis[i]){
      cnt=0;
      lm++;
      dfs(i);
      s[lm].sum=cnt;
    }
  sort(s+1,s+lm+1);
  bool flag=0;
  for(i=1;i<=lm;i++){
    if(s[i].sum&1)continue;
    else{
      if(s[i+1].sum==s[i].sum){i++;continue;}
      else {flag=1;break;}
    }
  }
  if(flag){printf("-1
");return 0;}
  for(i=1;i<=lm;i++){
    if(s[i].sum&1){
      for(j=0;j<s[i].sum;j++)
    q[j*2%s[i].sum]=s[i].p[j];
      for(j=0;j<s[i].sum-1;j++)
    ans[q[j]]=q[j+1];
      ans[q[s[i].sum-1]]=q[0];
    }
    else{
      int k=i+1;
      for(j=0;j<s[i].sum;j++){
    ans[s[i].p[j]]=s[k].p[j];
    ans[s[k].p[j]]=s[i].p[(j+1)%s[i].sum];
      }
      i++;
    }
  }
  for(i=1;i<=n;i++)
    printf("%d ",ans[i]);
  return 0;
}