1 #include <cstdio>
 2 #include <cmath>
 3 #include <algorithm>
 4 using namespace std;
 5 #define eps 1e-6
 6 #define N 25
 7 #define INF 20000
 8 #define pi acos(-1.0)
 9 struct point{
10     double x, y;
11     point(){}
12     point(double _x, double _y) {
13         x = _x, y = _y;
14     }
15 
16     point operator - (point a){
17         return point(x-a.x, y-a.y);
18     }
19 
20     double operator * (point a){
21         return x*a.y - y*a.x;
22     }
23 
24     double len(){
25         return sqrt(x*x+y*y);
26     }
27 };
28 struct circle{
29     point c;
30     double r;
31 };
32 circle cir[N];
33 int n;
34 
35 double dist(point a, point b)
36 {
37     return (a-b).len();
38 }
39 
40 double area_cir_to_cir(circle a,circle b)
41 {
42     double d=dist(a.c,b.c),r1=a.r,r2=b.r,r;
43     if (r1+r2<=d) { return 0.0; }
44     else if (fabs(r1-r2)>=d) {
45         r=min(r1,r2);
46         return pi*r*r;
47     }
48     else {
49         double a1=(r1*r1+d*d-r2*r2)/(2*r1*d);
50         double a2=(r2*r2+d*d-r1*r1)/(2*r2*d);
51         a1=2*acos(a1); a2=2*acos(a2);
52         return (r1*r1*(a1-sin(a1))+r2*r2*(a2-sin(a2)))*0.5;
53     }
54 }
55 
56 bool check(circle a, circle b)
57 {
58     double s1 = area_cir_to_cir(a, b);
59     double s2 = pi*b.r*b.r;
60     return s1*2 > s2-eps;
61 }//函数重载
62 
63 bool check(point o, double r)
64 {
65     circle t;
66     t.c = o, t.r = r;
67     for(int i = 0; i < n; i++)
68         if(!check(t, cir[i]))return false;
69     return true;
70 }
71 
72 double solve(int id)
73 {
74     point o = cir[id].c;
75     double l = 0, r = INF;
76     while(fabs(l-r) > eps)
77     {
78         double m = 0.5*(l+r);
79         if(check(o, m)) r = m;
80         else l = m;
81     }
82     return l;
83 }
84 
85 int main()
86 {
87     int T;
88     scanf("%d", &T);
89     while(T--)
90     {
91         scanf("%d", &n);
92         for(int i = 0; i < n; i++)
93             scanf("%lf %lf %lf", &cir[i].c.x, &cir[i].c.y, &cir[i].r);
94         double ans = INF;
95         for(int i = 0; i < n; i++)
96             ans = min(ans, solve(i));
97         printf("%.4f\n", ans);
98     }
99 }