【CodeForces 765D】 Artsem and Saunders（数学，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 print m 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 3
1 2 3

Input

3
2 2 2

Output

1
1 1 1
2

Input

2
2 1

Output

-1

题目大意：定义两个函数 g(h(x)) = x 和 h(g(x)) = f(x) 给出f(x) 构造 h(x) 和 g(x) 使其满足条件

思路：（我居然能写出D题，莫名兴奋）。g(x) 有 n 个元素， h(x) 有 m 个元素，根据函数关系，可以推出表达式 g(x) = g( f(x) ) ，此式表达 g(x)==g(a),其中a==f(x)，  设 g(1)=1，根据表达式从 1 ~n 构造出 g(x) ，构造过程中如果出现与之前矛盾的情况则输出 -1 。第二步根据  h(g(x)) = f(x) 从 1~n 推出 h(x) 的表达式，此过程中如果出现矛盾则输出 -1，完毕。

#include <bits/stdc++.h>
#define manx 100005
using namespace std;
int main()
{
    int f[manx]={0},h[manx]={0},g[manx]={0},n,m;

    scanf("%d",&n);
    for (int i=1; i<=n; i++) scanf("%d",&f[i]);
    int cot=1;
    for (int i=1; i<=n; i++){
        if (g[f[i]]) g[i]=g[f[i]];  //g[f[i]]在g[i]之前
        else if (g[i]==0) g[i]=cot++;
        if(!g[f[i]]) g[f[i]]=g[i]; //g[f[i]]在g[i]之后
        else if (g[f[i]]!=g[i]){
            printf("-1\n");
            return 0;
        }
    }
    m=0;
    for (int i=1; i<=n; i++){
        if(!h[g[i]]) {
            h[g[i]]=f[i];
            m++;
        }
        else if(h[g[i]]!=f[i]){
            printf("-1\n");
            return 0;
        }
    }
    printf("%d\n",m);
    for (int i=1; i<=n; i++)
        i==n ? printf("%d\n",g[i]) : printf("%d ",g[i]);
    for (int i=1; i<=m; i++)
        i==m ? printf("%d\n",h[i]) : printf("%d ",h[i]);
    return 0;
}