Square Root of Permutation - CF612E

Description

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.

Sample Input

 

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

简单题意

给出一个大小为n的置换群p，求一个置换群q，使得q^2=p

胡说题解

首先我们观察q^2是怎么运算的，置换群可以看成一个一个的环，每个点i向a[i]连一条边就是那个图

q^2其实就是把i的边变成a[a[i]]，也就是在环上走两步，然后q原本的环就变了

1.假设原来是奇数环，那么后来还是一个奇数环，只是顺序变了

2.假设原来是偶数环，那么就会拆成两个大小为一半的环

我们再看p

p上的奇数环可能是原来的奇数环，也有可能是偶数环拆开得来的

p上的偶数环只可能是原来的偶数环拆开得来

对于奇数环我们只要把这个环的后半部分与前半部分（先把这个环断开）交替插入就可以构造出原来的那个奇数环

对于偶数环我们就只能找一个相同大小的偶数环交替插入，即两个相同大小的偶数环合并，如果找不到相同大小的偶数环，那么我们就知道不存在这样的q使得q^2=p

#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn=1000100;

int n,tot,a[maxn],b[maxn],s[maxn],l[maxn],cir[maxn];
bool flag[maxn];

bool com(int a,int b){
    return l[a]<l[b];
}

int main(){
    scanf("%d",&n);
    int i,j,k;
    for(i=1;i<=n;i++)scanf("%d",&a[i]);
    for(i=1;i<=n;i++)
    if(!flag[i]){
        cir[++tot]=i;
        flag[i]=true;
        ++l[i];
        j=a[i];
        while(!flag[j]){
            flag[j]=true;
            ++l[i];
            j=a[j];
        }
    }
    sort(cir+1,cir+1+tot,com);
    int x=0;
    bool f=true;
    for(i=1;i<=tot;i++)
    if((l[cir[i]]&1)== 0){
        if(x==0)x=l[cir[i]];
        else
        if(x==l[cir[i]])x=0;
        else f=false;
    }
    if(x!=0)f=false;
    if(f==false)printf("-1");
    else{
        for(i=1;i<=tot;i++){
            if((l[cir[i]]&1)==1){
                j=cir[i];
                k=0;
                while(flag[j]){
                    s[k]=j;
                    flag[j]=false;
                    k=(k+2)%l[cir[i]];
                    j=a[j];
                }
                for(j=0;j<l[cir[i]]-1;j++)b[s[j]]=s[j+1];
                b[s[l[cir[i]]-1]]=s[0];
            }
            else{
                j=cir[i];
                k=cir[i+1];
                while(flag[j]){
                    b[j]=k;
                    b[k]=a[j];
                    flag[j]=false;
                    flag[k]=false;
                    j=a[j];
                    k=a[k];
                }
                ++i;
            }
        }
        for(i=1;i<=n;i++)printf("%d ",b[i]);
    }
    return 0;
}