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.

Sample Input:

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.

#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
int n, m, k, u, v, num, flag, ans;
vector<int>g[210];
int a[210];
bool f[210][210];
bool vis[210];

int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		f[i][i] = 1;
	}
	for (int i = 0; i < m; i++) {
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
		f[u][v] = f[v][u] = 1;
	}
	cin >> k;
	for (int t = 1; t <= k; t++) {
		cin >> num;
		flag = 0;
		memset(vis, 0, sizeof(vis));
		for (int i = 0; i < num; i++) {
			cin >> a[i];
			vis[a[i]] = 1;
		}
		for (int i = 0; i < num; i++) {
			if (flag == 1) {
				break;
			}
			for (int j = i + 1; j < num; j++) {
				if (!f[a[i]][a[j]]) {
					flag = 1;
					break;
				}
			}
		}
		if (flag == 1) {
			printf("Area %d needs help.\n", t);
		} else {
			for (int i = 1; i <= n; i++) {
				if (!vis[i]) {
					for (int j = 0; j < num; j++) {
						if (!f[i][a[j]]) {
							flag = 2;
							break;
						} else {
							ans = i;
							flag = 3;
						}
					}
				}
				if (flag == 3) {
					printf("Area %d may invite more people, such as %d.\n", t, ans);
					break;
				}
			}
			if (flag != 3) {
				printf("Area %d is OK.\n", t);
			}
		}
	}
	return 0;
}