Artsem has a friend Saunders from University of Chicago. Saunders PResented him with the following problem.

Let [n] denote the set {1,?...,?n}. We will also write f:?[x]?→?[y] when a function f is defined in integer points 1, ..., x, and all its values are integers from 1 to y.

Now then, you are given a function f:?[n]?→?[n]. Your task is to find a positive integer m, and two functions g:?[n]?→?[m],h:?[m]?→?[n], such that g(h(x))?=?x for all , and h(g(x))?=?f(x) for all , or determine that finding these is impossible.

The first line contains an integer n (1?≤?n?≤?105).

The second line contains n space-separated integers — values f(1),?...,?f(n) (1?≤?f(i)?≤?n).

If there is no answer, print one integer -1.

Otherwise, on the first line print the number m (1?≤?m?≤?106). On the second line print n numbers g(1),?...,?g(n). On the third line printm numbers h(1),?...,?h(m).

If there are several correct answers, you may output any of them. It is guaranteed that if a valid answer exists, then there is an answer satisfying the above restrictions.

31 2 3output
31 2 31 2 3input
32 2 2output
11 1 12input
22 1output
-1題解：
構造題。
因為： g(h(x))=x.h(g(x))=f(x).所以，h(x)=f(f(x)) && 必須有一個 x= f(x).(觀察可以發現)從而得到 g(x) =g(f(x)).
所以將這些x全部映射到[1,m]中，再映射到f(x)和g(x)中就可以了。
AC代碼：
#include<bits/stdc++.h>using namespace std;/*因為： g(h(x))=xh(g(x))=f(x)所以，h(x)=f(f(x)) && 必須有一個 x= f(x)從而得到 g(x) =g(f(x))*/int f[100000+7],g[100000+7];int h[100000+7];int main(){    int n;    int hn=0;    cin>>n;    for(int i=1;i<=n;i++){    	cin>>f[i];    	if(f[i]==i){ //x=f(x)     		hn++;    		g[i]=hn;    		h[g[i]]=i; //h(g(x))=f(x)=f(x);		}	}	//for(int i=1;i<=n;i++)cout<<"g[f[i]]="<<g[f[i]]<<endl;	for(int i=1;i<=n;i++){		if(!g[f[i]])return 0*puts("-1");	}	cout<<hn<<endl;      	                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       	for(int i=1;i<=n;i++)cout<<g[f[i]]<<" ";	cout<<endl;		for(int i=1;i<=hn;i++)	cout<<h[i]<<" ";	return 0;}