1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdio>
  4 #include<algorithm>
  5 #include<vector>
  6 #include<queue>
  7 #define N 105
  8 #define INF 2000000005
  9 using namespace std;
 10
 11 struct EDGE{
 12     int u, v, nt, cap;
 13     bool flag;
 14     bool vis;
 15     EDGE(){}
 16     EDGE(int u, int v, int nt, int cap, bool flag):u(u), v(v), nt(nt), cap(cap), flag(flag){}
 17 };
 18
 19 struct node{
 20     int x, y;
 21     node(){}
 22     node(int x, int y) : x(x), y(y){}
 23 };
 24
 25 int pos[10005];
 26
 27 node ans[10005];
 28 int preCost[20005];
 29 int vis[20005];
 30 int p[20005];
 31 int pcnt;
 32 int cnt;
 33
 34 vector<EDGE>g;
 35 int first[N];
 36
 37 int d[N];
 38 int n, e, c;
 39
 40 void addEdge(int u, int v, int c){
 41     g.push_back(EDGE(u, v, first[u], c, true));
 42     first[u] = g.size() - 1;
 43     g.push_back(EDGE(v, u, first[v], 0, false));
 44     first[v] = g.size() - 1;
 45 }
 46
 47 bool bfs(){
 48     memset(d, 0, sizeof(d));
 49     d[1] = 1;
 50     queue<int>q;
 51     q.push(1);
 52     while(!q.empty()){
 53         int u = q.front();
 54         q.pop();
 55         for(int i = first[u]; ~i; i = g[i].nt){
 56             int v = g[i].v;
 57             if(!d[v] && g[i].cap > 0){
 58                 d[v] = d[u] + 1;
 59                 q.push(v);
 60             }
 61         }
 62     }
 63     if(d[n] == 0) return false;
 64     return true;
 65 }
 66
 67 bool cmp(node a, node b){
 68     if(a.x == b.x)
 69          return a.y < b.y;
 70     return a.x < b.x;
 71 }
 72
 73 int leave;
 74
 75 int dfs(int u, int totf){
 76     int flow = 0;
 77     if(u ==n || totf==0) return totf;
 78     for(int i = first[u]; ~i; i = g[i].nt){
 79         int v = g[i].v;
 80         if(d[v] == d[u] + 1 && g[i].cap > 0){
 81             int ff  = dfs(v, min(totf-flow, g[i].cap));
 82             if(ff > 0){
 83                 if(!vis[i]){
 84                     p[pcnt++]=i;
 85                     preCost[i] = g[i].cap;
 86                     vis[i] = 1;
 87                 }
 88                 g[i].cap -= ff;
 89
 90                 if(!vis[i^1]){
 91                     p[pcnt++]=i^1;
 92                     preCost[i^1] = g[i^1].cap;
 93                     vis[i^1] = 1;
 94                 }
 95                 g[i^1].cap += ff;
 96                 flow += ff;
 97
 98                 if(flow >= leave){
 99                     flow = leave;
100                     return flow;
101                 }
102
103                 if(totf == flow) break;
104             }
105             else d[v] = -1;
106         }
107     }
108     return flow;
109 }
110
111 bool Dinic(){
112     leave = c;
113     while(bfs()){
114         leave -= dfs(1, INF);
115         if(leave == 0) break;
116     }
117     if(leave == 0) return true;
118     return false;
119 }
120
121
122
123 int main(){
124     int cas = 0;
125     while(scanf("%d%d%d", &n, &e, &c)){
126         if(!n) break;
127         memset(first, -1, sizeof(first));
128         g.clear();
129         cnt = 0;
130         while(e--){
131             int x, y, z;
132             scanf("%d%d%d", &x, &y, &z);
133             addEdge(x, y, z);
134         }
135         printf("Case %d: ", ++cas);//这一块差点没有把我气死...居然有一个空格，没有看清楚啊...一直PE.
136
137         if(n==1){
138             printf("possible\n");
139             continue;
140         }
141
142         if(Dinic())  printf("possible\n");
143         else{
144             int len = g.size();
145             for(int i=0; i<len; ++i)
146                 if(g[i].cap == 0 && g[i].flag)
147                     pos[cnt++] = i;//得到最小割
148             int cc = leave;//第一次Dinic之后，还剩下多少的流量需要流过
149             int ret = 0;
150             for(int i=0; i<cnt; ++i){
151                 c = cc;//新的需要流过的流量
152                 pcnt = 0;
153                 g[pos[i]].cap = INF;
154                 memset(vis, 0, sizeof(vis));
155                 if(Dinic())//如果增广成功，那么这条最小割满足
156                        ans[ret++] = node(g[pos[i]].u, g[pos[i]].v);
157                 for(int j=0; j<pcnt; ++j)
158                     g[p[j]].cap = preCost[p[j]];//将Dinic中所经过的边的值恢复成第一次Dinic之后的值！
159                 g[pos[i]].cap = 0;
160             }
161             if( ret > 0 ){
162                 sort(ans, ans+ret, cmp);
163                 printf("possible option:(%d,%d)", ans[0].x, ans[0].y);
164                 for(int i=1; i<ret; ++i)
165                     printf(",(%d,%d)", ans[i].x, ans[i].y);
166                 printf("\n");
167             }
168             else printf("not possible\n");
169         }
170     }
171     return 0;
172 }