Codeforces Round #397 D题Artsem and Saunders（构造法）解题报告

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 print mnumbers 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 3
1 2 3

input

3
2 2 2

output

1
1 1 1
2

input

2
2 1

output

-1

 

contest进行的时候一直卡在这道题。知道肯定是构造，但一直没想到合适的办法。结果比赛过后与同学交流，居然全都是用伪证构造过的。构造时只保证满足h（g(x))=f(x)这个条件，构造完成后检验另一个条件是否成立，如果成立则输出，不然就无解。（还没想到严格的证明orz）

#include <iostream>
#include<bits/stdc++.h>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int MAX=1e5+5;
int n;
int f[MAX],ff[MAX],g[MAX],h[MAX];
int m=0;
int main()
{
    scanf("%d",&n);
    int i;
    for(i=1;i<=n;i++)
    {
        scanf("%d",&f[i]);
    }
    for(i=1;i<=n;i++)
    {
        if(ff[f[i]]==0)
        {
            m++;
            h[m]=f[i];
            ff[f[i]]=m;
        }
        g[i]=ff[f[i]];
    }
    for(i=1;i<=m;i++)
    {
        if(g[h[i]]!=i)
        {
            printf("-1\n");
            return 0;
        }
    }
    printf("%d\n",m);
    for(i=1;i<=n;i++)
        printf("%d ",g[i]);
    puts("");
    for(i=1;i<=m;i++)
        printf("%d ",h[i]);

}