[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[],lm,ans[],q[];
 struct node{
   int sum;
   vector<int>p;
   friend bool operator < (const node a,const node b){
     return a.sum<b.sum;
   }
 }s[];
 int vis[],cnt;
 void dfs(int r){
   vis[r]=;
   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=;i<=n;i++)scanf("%d",&a[i]);
   for(i=;i<=n;i++)
     if(!vis[i]){
       cnt=;
       lm++;
       dfs(i);
       s[lm].sum=cnt;
     }
   sort(s+,s+lm+);
   bool flag=;
   for(i=;i<=lm;i++){
     if(s[i].sum&)continue;
     else{
       if(s[i+].sum==s[i].sum){i++;continue;}
       else {flag=;break;}
     }
   }
   if(flag){printf("-1\n");return ;}
   for(i=;i<=lm;i++){
     if(s[i].sum&){
       for(j=;j<s[i].sum;j++)
     q[j*%s[i].sum]=s[i].p[j];
       for(j=;j<s[i].sum-;j++)
     ans[q[j]]=q[j+];
       ans[q[s[i].sum-]]=q[];
     }
     else{
       int k=i+;
       for(j=;j<s[i].sum;j++){
     ans[s[i].p[j]]=s[k].p[j];
     ans[s[k].p[j]]=s[i].p[(j+)%s[i].sum];
       }
       i++;
     }
   }
   for(i=;i<=n;i++)
     printf("%d ",ans[i]);
   return ;
 }