A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0<N<100, the number of nodes in a tree, and M (<N), the number of non-leaf nodes. Then M lines follow, each in the format:
ID K ID[1] ID[2] … ID[K]
where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID’s of its children. For the sake of simplicity, let us fix the root ID to be 01.
The input ends with N being 0. That case must NOT be processed.

Output Specification:

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.
The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.

Sample Input:

2 1
01 1 02

Sample Output:

0 1

#include<iostream>
#include<map>
#include<vector>
using namespace std;
vector<int> v[100];
int book[100],maxdepth=-1;
void dfs(int index,int depth){
    if(v[index].size()==0){
        book[depth]++;
        maxdepth=max(maxdepth,depth);
        return;
    }
    for(int i=0;i<v[index].size();i++)
        dfs(v[index][i],depth+1);
}
int main()
{   
    int n,m,k,node,c;
    cin>>n>>m;
    for(int i=0;i<m;i++){
        cin>>node>>k;
        for(int j=0;j<k;j++){
            cin>>c;
            v[node].push_back(c);
        }
    }
    dfs(1,0);
    cout<<book[0];
    for(int i=1;i<=maxdepth;i++)
        cout<<" "<<book[i];
    return 0;
}
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