codefroces 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, ..., p_n (1 ≤ p_i ≤ 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 q_i (1 ≤ q_i ≤ 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

置换的整数幂有这样的结论：

T^k将长度为L的置换T分裂成gcd(L,K)份，每个循环分别是循环T中下标i mod gcd(l,k)=0,1,2…的元素的连接。

那么T^2分裂成gcd(L,2)分

也就是说，原置换平方后，偶数置换会分裂，奇数置换不变

开方运算实际上就是合并相同的置换

平方后的置换中偶数置换肯定是分裂后的结果，合并

奇数置换就不用合并

把循环求出来，排序

如果是偶数循环且没有长度相同的循环，那么说明无解，因为根本无法合并

注意奇数循环也要改变，因为奇数循环平方后顺序改变了

比如

1 4 2 5 3     ->1->4->2->5->3->

平方后就是

1 2 3 4 5     ->1->2->3->4->5->

给一张图直观理解(L=10,K=3)

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<vector>
 using namespace std;
 struct ZYYS
 {
   int sum;
   vector<int>p;
 }s[];
 int vis[],tot,a[],n,q[],ans[];
 bool cmp(ZYYS a,ZYYS b)
 {
   return a.sum<b.sum;
 }
 int gi()
 {
   char ch=getchar();
   int x=;
   while (ch<''||ch>'') ch=getchar();
   while (ch>=''&&ch<='')
     {
       x=x*+ch-'';
       ch=getchar();
     }
   return x;
 }
 int dfs(int x,int cnt)
 {
   if (vis[x]) return cnt;
   vis[x]=;
   s[tot].p.push_back(x);
   dfs(a[x],cnt+);
 }
 int main()
 {int i,flag=,j;
   cin>>n;
   for (i=;i<=n;i++)
     a[i]=gi();
   for (i=;i<=n;i++)
     if (vis[i]==)
     {
       s[++tot].sum=dfs(i,);
     }
   sort(s+,s+tot+,cmp);
   for (i=;i<=tot;i++)
     {
       if (s[i].sum&) continue;
       else
     {
       if (s[i+].sum==s[i].sum) {i++;continue;}
       else {flag=;break;}
     }
     }
   if (flag)
     {
       cout<<-<<endl;
       return ;
     }
   for (i=;i<=tot;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+)%s[i].sum];
     }
       else
     {
       for (j=;j<s[i].sum;j++)
         {
           ans[s[i].p[j]]=s[i+].p[j];
           ans[s[i+].p[j]]=s[i].p[(j+)%s[i].sum];
         }
       i++;
     }
     }
   for (i=;i<=n;i++)
     printf("%d ",ans[i]);
 }