/*
很经典的最大流最小割的题目
题意：求最小割，但因为最小割是不唯一的，题目要求得到最小割的条件下使得割边最少
搜到usaco类似的一个题目才出的，构造很巧妙
建边的时候每条边权 w=w*(E+1)+1;
这样得到最大流maxflow/(E+1) 就是答案了
道理很简单，如果原先两类割边都是最小割，那么求出的最大流相等
但边权变换后只有边数小的才是最小割了，至于为什么乘的是（E+1）是为了保证边数叠加后依然是余数，不至于影响求最小割的结果
*/
#include <cstdio>
#include <iostream>
#include <memory.h>
#include<queue>
using namespace std;
const int maxn=1009;
const __int64 inf=(1LL)<<60;
typedef __int64 LL;
//**************************************************
//为dinic求最大流模版
struct edge
{
int v, next;
LL val;
} net[ 500010 ];
int n,m;
int level[maxn], Qu[maxn], out[maxn],next[maxn];
class Dinic {
public:
int end;
Dinic() {
end = 0;
memset( next, -1, sizeof(next) );
}
inline void insert( int x, int y, LL c) {
net[end].v = y, net[end].val = c,
net[end].next = next[x],
next[x] = end ++;
net[end].v = x, net[end].val = 0,
net[end].next = next[y],
next[y] = end ++;
}
bool BFS( int S, int E ) {
memset( level, -1, sizeof(level) );
int low = 0, high = 1;
Qu[0] = S, level[S] = 0;
for( ; low < high; ) {
int x = Qu[low];
for( int i = next[x]; i != -1; i = net[i].next ) {
if( net[i].val == 0 ) continue;
int y = net[i].v;
if( level[y] == -1 ) {
level[y] = level[x] + 1;
Qu[ high ++] = y;
}
}
low ++;
}
return level[E] != -1;
}
LL MaxFlow( int S, int E ){
LL maxflow = 0;
for( ; BFS(S, E) ; ) {
memcpy( out, next, sizeof(out) );
int now = -1;
for( ;; ) {
if( now < 0 ) {
int cur = out[S];
for(; cur != -1 ; cur = net[cur].next )
if( net[cur].val && out[net[cur].v] != -1 && level[net[cur].v] == 1 )
break;
if( cur >= 0 ) Qu[ ++now ] = cur, out[S] = net[cur].next;
else break;
}
int u = net[ Qu[now] ].v;
if( u == E ) {
LL flow = inf;
int index = -1;
for( int i = 0; i <= now; i ++ ) {
if( flow > net[ Qu[i] ].val )
flow = net[ Qu[i] ].val, index = i;
}
maxflow += flow;
for( int i = 0; i <= now; i ++ )
net[Qu[i]].val -= flow, net[Qu[i]^1].val += flow;
for( int i = 0; i <= now; i ++ ) {
if( net[ Qu[i] ].val == 0 ) {
now = index - 1;
break;
}
}
}
else{
int cur = out[u];
for(; cur != -1; cur = net[cur].next )
if (net[cur].val && out[net[cur].v] != -1 && level[u] + 1 == level[net[cur].v])
break;
if( cur != -1 )
Qu[++ now] = cur, out[u] = net[cur].next;
else out[u] = -1, now --;
}
}
}
return maxflow;
}
};
int main()
{
int ca,t,u,v;
LL w;
scanf("%d",&ca);
for(int k=1;k<=ca;k++)
{
Dinic my;
scanf("%d%d",&n,&m);
while(m--)
{
scanf("%d%d%I64d%d",&u,&v,&w,&t);
my.insert(u,v,w*100001+1);
if(t==1)
{
my.insert(v,u,w*100001+1);
}
}
LL ans=my.MaxFlow(0,n-1);
//cout<<ans<<endl;
printf("Case %d: %I64d\n",k,(ans%100001));
}
return 0;
}
