A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a list of schools to which it distributes software (the “receiving schools”). Note that if B is in the distribution list of school A, then A does not necessarily appear in the list of school B
You are to write a program that computes the minimal number of schools that must receive a copy of the new software in order for the software to reach all schools in the network according to the agreement (Subtask A). As a further task, we want to ensure that by sending the copy of new software to an arbitrary school, this software will reach all schools in the network. To achieve this goal we may have to extend the lists of receivers by new members. Compute the minimal number of extensions that have to be made so that whatever school we send the new software to, it will reach all other schools (Subtask B). One extension means introducing one new member into the list of receivers of one school.

Input

The first line contains an integer N: the number of schools in the network (2 <= N <= 100). The schools are identified by the first N positive integers. Each of the next N lines describes a list of receivers. The line i+1 contains the identifiers of the receivers of school i. Each list ends with a 0. An empty list contains a 0 alone in the line.

Output

Your program should write two lines to the standard output. The first line should contain one positive integer: the solution of subtask A. The second line should contain the solution of subtask B.

Sample Input

5
2 4 3 0
4 5 0
0
0
1 0

Sample Output

1
2

題意：給出n個點的有向圖，問1.最少要多少個點才能遍歷全部邊，2.最少要加多少條邊才能使其成為連通圖。

思路：強連通縮點，出度個數就是需要的最少遍歷圖的點個數，max（入度，出度）為最少增加邊數，可使成為連通圖。

#include<algorithm>
#include<string.h>
#include<stdio.h>
#include<stack>
using namespace std;
struct node
{
    int to,nextt;
}A[10010];
int tot,head[110],carry,indox;
int DFN[110],LOW[110],book[110],ans[110];
int in[110],out[110];
stack<int>q;
void add(int x,int y)
{
    ++tot;
    A[tot].to=y;
    A[tot].nextt=head[x];
    head[x]=tot;
}
int init()
{
    tot=carry=indox=0;
    memset(head,0,sizeof(head));
    memset(DFN,-1,sizeof(DFN));
    memset(LOW,-1,sizeof(LOW));
    memset(book,0,sizeof(book));
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
}
int tarjan(int x)
{
    int tem;
    DFN[x]=LOW[x]=++carry;
    book[x]=1;
    q.push(x);
    for(int i=head[x];i!=0;i=A[i].nextt)
    {
        tem=A[i].to;
        if(DFN[tem]==-1)
        {
            tarjan(tem);
            LOW[x]=min(LOW[x],LOW[tem]);
        }
        else if(book[tem]==1)
        {
            LOW[x]=min(LOW[x],DFN[tem]);
        }
    }
    if(LOW[x]==DFN[x])
    {
        ++indox;
        do
        {
            tem=q.top();
            book[tem]=0;
            ans[tem]=indox;
            q.pop();
        }while(tem!=x);
    }
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        init();
        int x;
        for(int i=1;i<=n;i++)
        {
            while(~scanf("%d",&x)&&x)
            {
                add(i,x);
            }
        }
        for(int i=1;i<=n;i++)
        {
            if(DFN[i]==-1)
                tarjan(i);
        }
        if(indox==1)
        {
            printf("1\n0\n");
            continue;
        }
        int ans1=0,ans2=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=head[i];j!=0;j=A[j].nextt)
            {
                if(ans[i]!=ans[A[j].to])
                {
                    in[ans[A[j].to]]=1;
                    out[ans[i]]=1;
                }
            }
        }
        for(int i=1;i<=indox;i++)
        {
            if(in[i]==0)
                ans1++;
            if(out[i]==0)
                ans2++;
        }
        printf("%d\n%d\n",ans1,max(ans1,ans2));
    }
}