Codeforces Round #397 D. Artsem and Saunders（构造）

D. Artsem and Saunders

time limit per test
2 seconds

memory limit per test
512 megabytes

input
standard input

output
standard output

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.

Input

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).

Output

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.

Examples

input

3
1 2 3

output

3
1 2 31 2 3

input

3
2 2 2

output

1
1 1 1
2

input

2
2 1

output

-1

题解：

构造题。

因为： 

g(h(x))=x.

h(g(x))=f(x).

所以，h(x)=f(f(x)) && 必须有一个 x= f(x).(观察可以发现)

从而得到 g(x) =g(f(x)).

所以将这些x全部映射到[1,m]中，再映射到f(x)和g(x)中就可以了。

AC代码：

#include<bits/stdc++.h>
using namespace std;
/*
因为：
g(h(x))=x
h(g(x))=f(x)
所以，h(x)=f(f(x)) && 必须有一个 x= f(x)
从而得到 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;
}