"Damn Single (单身狗)" is the Chinese nickname for someone who is being single. You are supposed to find those who are alone in a big party, so they can be taken care of.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=50000), the total number of couples. Then N lines of the couples follow, each gives a couple of ID's which are 5-digit numbers (i.e. from 00000 to 99999). After the list of couples, there is a positive integer M (<=10000) followed by M ID's of the party guests. The numbers are separated by spaces. It is guaranteed that nobody is having bigamous marriage (重婚) or dangling with more than one companion.

Output Specification:

First print in a line the total number of lonely guests. Then in the next line, print their ID's in increasing order. The numbers must be separated by exactly 1 space, and there must be no extra space at the end of the line.

Sample Input:

3

11111 22222

33333 44444

55555 66666

7

55555 44444 10000 88888 22222 11111 23333

Sample Output:

5

10000 23333 44444 55555 88888


标记每一对情侣，然后查看有情侣的，情侣是否在场就好。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int love[],haslove[],loveex[],a,b,c;

int ans[],peo[],n,m;

int main()

{

    cin>>n;

    for(int i = ;i < n;i ++)

    {

        cin>>a>>b;

        love[a] = b;

        love[b] = a;

        haslove[a] = ;

        haslove[b] = ;

    }

    cin>>m;

    for(int i = ;i < m;i ++)

    {

        cin>>peo[i];

        if(haslove[peo[i]])loveex[love[peo[i]]] = ;

    }

    for(int i = ;i < m;i ++)

    {

        if(!loveex[peo[i]])ans[c ++] = peo[i];

    }

    sort(ans,ans + c);

    cout<<c<<endl;

    if(c)printf("%05d",ans[]);

    for(int i = ;i < c;i ++)

        printf(" %05d",ans[i]);

}