1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstdlib>
  4 #include <cstring>
  5 #include <algorithm>
  6 #include <vector>
  7 
  8 using namespace std;
  9 
 10 #define N 19
 11 #define ll long long
 12 
 13 ll ma;
 14 ll a[N];
 15 int p[N];
 16 int have[N];
 17 int n;
 18 ll dp[ (1 << 17) ][N];
 19 int tot;
 20 ll inf;
 21 vector<int> G[N];
 22 
 23 int cal(int x)
 24 {
 25     int ret = 0;
 26     while(x){
 27         ret += (x & 1);
 28         x /= 2;
 29     }
 30     return ret;
 31 }
 32 
 33 void ini()
 34 {
 35     int o;
 36     for(o = 0;o < n;o++){
 37         G[o].clear();
 38     }
 39     for(o = 0;o < tot;o++){
 40         G[ cal(o) ].push_back(o);
 41     }
 42 }
 43 
 44 int main()
 45 {
 46 
 47     int T;
 48     //freopen("in.txt","r",stdin);
 49     //freopen("out.txt","w",stdout);
 50     scanf("%d",&T);
 51     for(int ccnt = 1;ccnt <= T;ccnt++){
 52         scanf("%d",&n);
 53         inf = (1LL << 60);
 54         ma = -inf;
 55         tot = (1 << n);
 56         ini();
 57         int i;
 58         memset(have,-1,sizeof(have));
 59         for(i = 0;i < n;i++){
 60             scanf("%I64d%d",&a[i],&p[i]);
 61             if(p[i] != -1){
 62                 have[ p[i] ] = i;
 63             }
 64         }
 65         int o,nt;
 66         for(o = 0;o < tot;o++){
 67             for(i = 0;i < n;i++){
 68                 dp[o][i] = -inf;
 69             }
 70         }
 71         for(i = 0;i < n;i++){
 72             if(p[i] == 0 || p[i] == -1){    //放在第一个
 73                 dp[ (1 << i) ][i] = 0;
 74             }
 75         }
 76 
 77         /*
 78         for(i = 0;i < n;i++){
 79             for(int j = 0;j < G[i].size();j++){
 80                 printf(" i =%d j = %d g= %d\n",i,j,G[i][j]);
 81             }
 82         }*/
 83         for(i = 1;i < n;i++){      //放第i位
 84             int sz = G[i].size();
 85             for(int j = 0;j < sz;j++){  //遍历含有i个1的所有数
 86                 o = G[i][j];
 87                 for(int k = 0;k < n;k++)    //把第k个数放在第i位
 88                 {
 89 
 90                     if( o & (1 << k) ) continue;    //k已经放了
 91                     if( p[k] != -1 && p[k] != i ) continue; //k被固定了
 92                     if( have[i] != -1 && have[i] != k ) continue;   //位置i被固定了
 93                     nt = o | (1 << k);
 94                     //printf(" i = %d o = %d k = %d nt = %d\n",i,o,k,nt);
 95                     for(int pr = 0;pr < n;pr++){
 96                         if( (o & (1 << pr ) ) == 0) continue;    //pr不在o里
 97                         if( dp[o][pr] == -inf ) continue;
 98                         //printf(" i = %d o = %d k = %d nt = %d pr = %d\n",i,o,k,nt,pr);
 99                         dp[nt][k] = max(dp[nt][k],dp[o][pr] + a[pr] * a[k]);
100                     }
101                 }
102             }
103         }
104         /*
105         for(o = 0;o < tot;o++){
106             for(i = 0;i < n;i++){
107                 printf(" o = %d i = %d dp = %I64d\n",o,i,dp[o][i]);
108             }
109         }*/
110         for(i = 0;i < n;i++){
111             ma = max(ma,dp[tot - 1][i]);
112         }
113         printf("Case #%d:\n",ccnt);
114         printf("%I64d\n",ma);
115     }
116 
117     return 0;
118 }