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[ C14>i] = q[q[i]] for each i = 1 ... n. For example, the square of q = [4, 5, 1, 2, 3] is p = Q^{2 = [ 2, 3, 4, 5, 1].}

This problem are about the inverse operation:given the permutation p You task are to find such permutation Q that q^{2 = p. If There is several such Q find any of them.}

Input

The first line contains integer n (1≤ n ≤10^{6)-the number of elements in Permutati On P.}

The second line contains n distinct integers p_{1, P2, ..., pn (1≤ pi ≤ n)-the elements of permutation p.}

Output

If there is no permutation q such that q^{2 = P Print the number "-1". /c23>}

If The answer exists print it. The only line should contain n different integers q_{i (1≤ qi< /c9>≤ n)-the elements of the permutation q. If There is 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

Simple test Instructions

A permutation group p with a size of n is given, and a permutation group Q is obtained, which makes the q^2=p

Nonsense

First we observe how q^2 is calculated, the permutation group can be regarded as a ring, each point I to a[i] even a side is that figure

Q^2 actually is to turn the side of I into a[a[i], that is, two steps on the ring, then q the original ring has changed

1. Assuming the original is an odd ring, then it is an odd ring, but the order has changed

2. If the original is an even ring, then it will be split into two half-size rings

We'll see P again.

The odd ring on p may be the original odd ring, or it may be an even ring apart

The even ring on p can only be disassembled by the original even ring.

For odd rings We can construct the original odd ring by alternately inserting the back part of the ring with the first half (breaking the ring first).

For even rings we can only find an even ring of the same size alternately, that is, two even rings of the same size are merged, and if we cannot find even rings of the same size, then we know that there is no such q which makes q^2=p

1#include <cstdio>2#include <algorithm>3 using namespacestd;4 5 Const intmaxn=1000100;6 7 intN,TOT,A[MAXN],B[MAXN],S[MAXN],L[MAXN],CIR[MAXN];8 BOOLFLAG[MAXN];9 Ten BOOLcomintAintb) { One     returnl[a]<L[b]; A } -  - intMain () { thescanf"%d",&n); -     inti,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; A++L[i]; atj=A[i]; -          while(!Flag[j]) { -flag[j]=true; -++L[i]; -j=A[j]; -         } in     } -Sort (cir+1, cir+1+tot,com); to     intx=0; +     BOOLf=true; -      for(i=1; i<=tot;i++) the     if((l[cir[i]]&1)==0){ *         if(x==0) x=L[cir[i]]; $         ElsePanax Notoginseng         if(X==l[cir[i]]) x=0; -         Elsef=false; the     } +     if(x!=0) f=false; A     if(f==false) printf ("-1"); the     Else{ +          for(i=1; i<=tot;i++){ -             if((l[cir[i]]&1)==1){ $j=Cir[i]; $k=0; -                  while(Flag[j]) { -s[k]=J; theflag[j]=false; -k= (k +2)%L[cir[i]];Wuyij=A[j]; the                 } -                  for(j=0; j<l[cir[i]]-1; j + +) b[s[j]]=s[j+1]; Wub[s[l[cir[i]]-1]]=s[0]; -             } About             Else{ $j=Cir[i]; -k=cir[i+1]; -                  while(Flag[j]) { -b[j]=K; Ab[k]=A[j]; +flag[j]=false; theflag[k]=false; -j=A[j]; $k=A[k]; the                 } the++i; the             } the         } -          for(i=1; i<=n;i++) printf ("%d", B[i]); in     } the     return 0; the}

AC Code

Square Root of permutation-cf612e