A summit (峰会) is a meeting of heads of state or government. Arranging the rest areas for the summit is not a simple job. The ideal arrangement of one area is to invite those heads so that everyone is a direct friend of everyone.

Now given a set of tentative arrangements, your job is to tell the organizers whether or not each area is all set.

## Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N (≤ 200), the number of heads in the summit, and M, the number of friendship relations. Then M lines follow, each gives a pair of indices of the heads who are friends to each other. The heads are indexed from 1 to N.

Then there is another positive integer K (≤ 100), and K lines of tentative arrangement of rest areas follow, each first gives a positive number L (≤ N), then followed by a sequence of L distinct indices of the heads. All the numbers in a line are separated by a space.

## Output Specification:

For each of the K areas, print in a line your advice in the following format:

if in this area everyone is a direct friend of everyone, and no friend is missing (that is, no one else is a direct friend of everyone in this area), print Area X is OK..

if in this area everyone is a direct friend of everyone, yet there are some other heads who may also be invited without breaking the ideal arrangement, print Area X may invite more people, such as H. where H is the smallest index of the head who may be invited.

if in this area the arrangement is not an ideal one, then print Area X needs help. so the host can provide some special service to help the heads get to know each other.

Here X is the index of an area, starting from 1 to K.

8 10
5 6
7 8
6 4
3 6
4 5
2 3
8 2
2 7
5 3
3 4
6
4 5 4 3 6
3 2 8 7
2 2 3
1 1
2 4 6
3 3 2 1

## Sample Output:

Area 1 is OK.
Area 2 is OK.
Area 3 is OK.
Area 4 is OK.
Area 5 may invite more people, such as 3.
Area 6 needs help.

## 解答

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

using namespace std;

int parent[210];
int Find(int a) {
return a == parent[a] ? a : parent[a] = Find(parent[a]);
}
void Union(int a, int b) {
int pa = Find(a);
int pb = Find(b);
if (pa != pb) parent[pb] = pa;
}
bool check(vector<vector<bool>> &isFriend, vector<int> heads) {
for (int i = 0; i < heads.size(); i++) {
for (int j = i + 1; j < heads.size(); j++)
}
return true;
}

int main() {
int n, m, a, b;
scanf("%d %d", &n, &m);
vector<vector<bool>> isFriend(n + 1, vector<bool>(n + 1, false));
for (int i = 0; i < 211; i++) parent[i] = i;
for (int i = 0; i < m; i++) {
scanf("%d %d", &a, &b);
isFriend[a][b] = true;
isFriend[b][a] = true;
Union(a, b);
}
int testn, num;
scanf("%d", &testn);
for (int j = 0; j < testn; j++) {
scanf("%d", &num);
unordered_map<int, bool> mp;
for (int k = 0; k < num; k++) {
}
printf("Area %d needs help.\n", j + 1);
continue;
}
int k;
bool flag = true;
for (k = 1; k <= n; k++) {
if (mp.find(k) != mp.end()) continue;
int t = 0;
for (; t < num && isFriend[heads[t]][k]; t++);
if (t == num) {
flag = false;
break;
}
}
if (!flag) printf("Area %d may invite more people, such as %d.\n", j + 1, k);
else printf("Area %d is OK.\n", j + 1);
}
return 0;
}