1 /********************
  2 User：Mandy
  3 Language:c++
  4 Problem:
  5 ********************/
  6 #include<bits/stdc++.h>
  7
  8 using namespace std;
  9
 10 typedef long long ll;
 11 typedef double dou;
 12
 13 const int maxn = 1005;
 14 const dou esp = 1e-8;
 15
 16 int t,n,size;
 17 double d[maxn];
 18 int from[maxn];
 19 double dis[maxn][maxn],cost[maxn][maxn];
 20
 21 struct Node{
 22     int x,y,z;
 23 }node[maxn];
 24
 25
 26 template<class T>inline void read(T &x){
 27     x = 0;bool flag = 0;char ch = getchar();
 28     while(!isdigit(ch)) flag |= ch == '-',ch = getchar();
 29     while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48),ch = getchar();
 30     if(flag) x = -x;
 31 }
 32
 33 template<class T>void putch(const T x){
 34     if(x > 9) putch(x / 10);
 35     putchar(x % 10 | 48);
 36 }
 37
 38 template<class T>void put(const T x){
 39     if(x < 0) putchar('-'),putch(-x);
 40     else putch(x);
 41 }
 42
 43 void file(){
 44     freopen("build3.in","r",stdin);
 45     freopen("build.out","w",stdout);
 46 }
 47
 48 void readdata(){
 49     for(int i = 1;i <= n; ++ i){
 50         read(node[i].x);
 51         read(node[i].y);
 52         read(node[i].z);
 53     }
 54 }
 55
 56 double dist(int i,int j){
 57     return sqrt(((dou)node[i].x - node[j].x) *
 58                 ((dou)node[i].x - node[j].x) +
 59                 ((dou)node[i].y - node[j].y) *
 60                 ((dou)node[i].y - node[j].y));
 61 }
 62
 63 void init(){
 64     for(int i = 1;i <= n; ++ i){
 65         for(int j = i;j <= n; ++ j){
 66             dis[i][j] = dist(i,j);
 67             dis[j][i] = dis[i][j];
 68             cost[i][j] = abs((double)node[i].z - node[j].z);
 69             cost[j][i] = cost[i][j];
 70         }
 71     }
 72 }
 73
 74 double Prim(double r){
 75     int cnt = 1;t = 0;double c = 0,D = 0;
 76     for(int i = 1;i <= n;++i)
 77         d[i] = cost[1][i]-r*dis[1][i],from[i] = 1;
 78     d[1] = -2e9;
 79     for(int j = 1;j <= n; ++ j){
 80         double cur = 2e9;int id = 0;
 81         for(int i = 1;i <= n; ++ i){
 82             if(d[i]!=-2e9 && d[i] < cur){
 83                 cur = d[i];    id = i;
 84             }
 85         }
 86         t += cur;cnt++;d[id] = -2e9;
 87         c += cost[from[id]][id];
 88         D += dis[from[id]][id];
 89         if(cnt == n) break;
 90         for(int i = 1;i <= n; ++ i) {
 91             if(d[i] > cost[id][i]-r*dis[id][i]){
 92                 d[i] = cost[id][i]-r*dis[id][i];
 93                 from[i] = id;
 94             }
 95         }
 96     }
 97     if(t < esp) return c/D;
 98     else return -2e9;
 99 }
100
101 void work(){
102     double ans = (double)7;
103     while(1){
104         double R = Prim(ans);
105         if(R == -2e9||fabs(ans - R) <= esp) break;//fabs!!
106         //直接判断是否相等就好，r不会等于-2e9
107         ans = R;
108     }
109     printf("%.3lf\n",ans);
110 }
111
112 int main(){
113 //    file();
114     while(~scanf("%d",&n)){
115         if(!n) break;
116         size = 0;
117         readdata();
118         init();
119         work();
120     }
121     return 0;
122 }