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[1000001];
int vis[1000001],tot,a[1000001],n,q[1000001],ans[1000001];
bool cmp(ZYYS a,ZYYS b)
{
  return a.sum<b.sum;
}
int gi()
{
  char ch=getchar();
  int x=0;
  while (ch<'0'||ch>'9') ch=getchar();
  while (ch>='0'&&ch<='9')
    {
      x=x*10+ch-'0';
      ch=getchar();
    }
  return x;
}
int dfs(int x,int cnt)
{
  if (vis[x]) return cnt;
  vis[x]=1;
  s[tot].p.push_back(x);
  dfs(a[x],cnt+1);
}
int main()
{int i,flag=0,j;
  cin>>n;
  for (i=1;i<=n;i++)
    a[i]=gi();
  for (i=1;i<=n;i++)
    if (vis[i]==0)
    {
      s[++tot].sum=dfs(i,0);
    }
  sort(s+1,s+tot+1,cmp);
  for (i=1;i<=tot;i++)
    {
      if (s[i].sum&1) continue;
      else
    {
      if (s[i+1].sum==s[i].sum) {i++;continue;}
      else {flag=1;break;}
    }
    }
  if (flag)
    {
      cout<<-1<<endl;
      return 0;
    }
  for (i=1;i<=tot;i++)
    {
      if (s[i].sum&1)
    {
      for (j=0;j<s[i].sum;j++)
        {
          q[(2*j)%s[i].sum]=s[i].p[j];
        }
      for (j=0;j<s[i].sum;j++)
        ans[q[j]]=q[(j+1)%s[i].sum];
    }
      else
    {
      for (j=0;j<s[i].sum;j++)
        {
          ans[s[i].p[j]]=s[i+1].p[j];
          ans[s[i+1].p[j]]=s[i].p[(j+1)%s[i].sum];
        }
      i++;
    }
    }
  for (i=1;i<=n;i++)
    printf("%d ",ans[i]);
}