Tom is a commander, his task is destroying his enemy’s transportation system.

Let’s represent his enemy’s transportation system as a simple directed graph G with n nodes and m edges. Each node is a city and each directed edge is a directed road. Each edge from node u to node v is associated with two values D and B, D is the cost to destroy/remove such edge, B is the cost to build an undirected edge between u and v.

His enemy can deliver supplies from city u to city v if and only if there is a directed path from u to v. At first they can deliver supplies from any city to any other cities. So the graph is a strongly-connected graph.

He will choose a non-empty proper subset of cities, let’s denote this set as S. Let’s denote the complement set of S as T. He will command his soldiers to destroy all the edges (u, v) that u belongs to set S and v belongs to set T. 

To destroy an edge, he must pay the related cost D. The total cost he will pay is X. You can use this formula to calculate X:

After that, all the edges from S to T are destroyed. In order to deliver huge number of supplies from S to T, his enemy will change all the remained directed edges (u, v) that u belongs to set T and v belongs to set S into undirected edges. (Surely, those edges exist because the original graph is strongly-connected)

To change an edge, they must remove the original directed edge at first, whose cost is D, then they have to build a new undirected edge, whose cost is B. The total cost they will pay is Y. You can use this formula to calculate Y:

At last, if Y>=X, Tom will achieve his goal. But Tom is so lazy that he is unwilling to take a cup of time to choose a set S to make Y>=X, he hope to choose set S randomly! So he asks you if there is a set S, such that Y<X. If such set exists, he will feel unhappy, because he must choose set S carefully, otherwise he will become very happy.

There are multiply test cases.

The first line contains an integer T(T<=200), indicates the number of cases. 

For each test case, the first line has two numbers n and m. 

Next m lines describe each edge. Each line has four numbers u, v, D, B. 
(2=<n<=200, 2=<m<=5000, 1=<u, v<=n, 0=<D, B<=100000)

The meaning of all characters are described above. It is guaranteed that the input graph is strongly-connected.

For each case, output "Case #X: " first, X is the case number starting from 1.If such set doesn’t exist, print “happy”, else print “unhappy”.

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-10
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

const int maxn = 2010;

using namespace std;

struct node
{
    int x, y;
} f[maxn];

int main()
{
    int T;
    cin >>T;
    int Case = 1;
    while(T--)
    {
        int n, m;
        cin >>n>>m;
        int x, y;
        int d, b;
        for(int i = 0; i <= n; i++) f[i].x = f[i].y = 0;
        for(int i = 0; i < m; i++)
        {
            scanf("%d %d %d %d",&x, &y, &d, &b);
            f[x].y += d;
            f[y].x += (d+b);
        }
        int flag = 0;
        for(int i = 1; i <= n; i++)
        {
            if(f[i].x < f[i].y)
            {
                flag = 1;
                break;
            }
        }
        cout<<"Case #"<<Case++<<": ";
        if(flag) cout<<"unhappy"<<endl;
        else cout<<"happy"<<endl;
    }
    return 0;
}