"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 (≤ 50,000), 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 (≤ 10,000) 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

## 解答

#include<iostream>
#include<unordered_map>
#include<vector>
#include<algorithm>

using namespace std;

int main()
{
int cn, gn, a, b;
cin >> cn;
unordered_map<int, int> mp;
vector<int> ans;
while (cn--) {
cin >> a >> b;
mp[a] = b;
mp[b] = a;
}
cin >> gn;
unordered_map<int, bool> guest;
while (gn--) {
cin >> a;
guest[a] = true;
}
for (auto it = guest.begin(); it != guest.end(); it++) {
if (mp.find(it->first) == mp.end() ||
(mp.find(it->first) != mp.end() &&
guest.find(mp[it->first]) == guest.end()))
ans.push_back(it->first);
}
printf("%d\n", ans.size());
if (ans.size() > 0) {
sort(ans.begin(), ans.end());
for (auto it = ans.begin(); it != ans.end() - 1; it++)
printf("%05d ", *it);
printf("%05d", ans.back());
}
return 0;
}