#include <cstdio>

 #include <cmath>

 #include <cstring>

 #include <algorithm>

 #include <vector>

 using namespace std;

 const int maxn =  + ;

 const double eps = 1e-;        //别开太大，样例数据就到达1e-11级别

 const double pi = acos(-);

 int dcmp(double x)

 {

     return fabs(x) < eps ?  : (x >  ?  : -);

 }

 struct Point

 {

     double x;

     double y;

     Point(double x = , double y = ):x(x), y(y) {}

     bool operator < (const Point& e) const

     {

         return dcmp(x - e.x) <  || (dcmp(x - e.x) ==  && dcmp(y - e.y) < );

     }

     bool operator == (const Point& e) const

     {

         return dcmp(x - e.x) ==  && dcmp(y - e.y) == ;

     }

     int read()

     {

         return scanf("%lf%lf", &x, &y);

     }

 };

 typedef Point Vector;

 Vector operator + (Point A, Point B)

 {

     return Vector(A.x + B.x, A.y + B.y);

 }

 Vector operator - (Point A, Point B)

 {

     return Vector(A.x - B.x, A.y - B.y);

 }

 Vector operator * (Point A, double p)

 {

     return Vector(A.x * p, A.y * p);

 }

 Vector operator / (Point A, double p)

 {

     return Vector(A.x / p, A.y / p);

 }

 struct Circle

 {

     Point c;

     double r;

     Circle() {}

     Circle(Point c, double r):c(c), r(r) {}

     int read()

     {

         return scanf("%lf%lf%lf", &c.x, &c.y, &r);

     }

     Point point(double a)

     {

         return Point(c.x + r * cos(a), c.y + r * sin(a));

     }

 };

 double Dot(Vector A, Vector B)

 {

     return A.x * B.x + A.y * B.y;

 }

 double Length(Vector A)

 {

     return sqrt(Dot(A, A));

 }

 double angle(Vector v)      //求向量的极角

 {

     return atan2(v.y, v.x);

 }

 bool PointInCircle(Point p, Circle C)       //判断点是否在圆内

 {

     double dist = Length(p - C.c);

     if(dcmp(dist - C.r) > ) return ;      //这里我选择点在圆边上不算在圆内

     else return ;

 }

 bool CircleInCircle(Circle A, Circle B)         //判断圆在圆内

 {

     double cdist = Length(A.c - B.c);

     double rdiff = B.r - A.r;

     if(dcmp(A.r - B.r) <=  && dcmp(cdist - rdiff) <= ) return ;      //包括重合，内切和内含的情况

     return ;

 }

 int n;

 Circle C[maxn];

 bool vis[maxn];

 vector<double> pointAng[maxn];

 int GetCircleCircleIntersection(int c1, int c2)       //求圆与圆的交点

 {

     Circle C1 = C[c1];

     Circle C2 = C[c2];

     double d = Length(C1.c - C2.c);

     if(dcmp(d) == )

     {

         if(dcmp(C1.r - C2.r) == ) return -;       //两圆重合

         return ;       //同心圆但不重合

     }

     if(dcmp(C1.r + C2.r - d) < ) return ;     //外离

     if(dcmp(fabs(C1.r - C2.r) - d) > ) return ;       //内含

     double a = angle(C2.c - C1.c);

     double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / ( * C1.r * d));

     Point p1 = C1.point(a + da);

     Point p2 = C1.point(a - da);

     if(p1 == p2) return ;      //相切

     pointAng[c1].push_back(a + da);     //相切的点不处理，只要相交的

     pointAng[c1].push_back(a - da);

     return ;

 }

 void init()

 {

     for(int i = ; i < n; i++) pointAng[i].clear();

     memset(vis, , sizeof(vis));

 }

 void read()

 {

     for(int i = ; i < n; i++) C[i].read();

 }

 void solve()

 {

     for(int i = ; i < n; i++)      //圆两两相交，得各圆交点集合

         for(int j = ; j < n; j++) if(i != j)

                 GetCircleCircleIntersection(i, j);

     for(int i = ; i < n; i++)

     {

         sort(pointAng[i].begin(), pointAng[i].end());       //各圆交点按极角排序

         vector<double>::iterator iter = unique(pointAng[i].begin(), pointAng[i].end());     //去重，可减少运行时间，不去重也能AC

         pointAng[i].resize(distance(pointAng[i].begin(), iter));

     }

     for(int i = ; i < n; i++)      //判断第i个圆上的弧

     {

         int sz = pointAng[i].size();

         if(!sz)         //此圆不与其他圆相交

         {

             bool ok = ;

             for(int k = i+; k < n; k++) if(CircleInCircle(C[i], C[k]))         //判上面是否有圆把它覆盖掉

                 {

                     ok = ;

                     break;

                 }

             if(ok) vis[i] = ;

         }

         else

         {

             pointAng[i].push_back(pointAng[i][]);

             for(int j = ; j < sz; j++)         //第i个圆上的第j条弧

             {

                 bool ok = ;

                 Point pm = C[i].point((pointAng[i][j] + pointAng[i][j+]) / );     //取弧的中点

                 for(int k = i+; k < n; k++) if(PointInCircle(pm, C[k]))

                     {

                         ok = ;

                         break;

                     }

                 if(ok)

                 {

                     vis[i] = ;

                     for(int u = i-; u >= ; u--)if(PointInCircle(pm, C[u]))        //把这段圆弧下的圆设为可见

                         {

                             vis[u] = ;

                             break;

                         }

                 }

             }

         }

     }

     int ret = ;

     for(int i = ; i < n; i++) if(vis[i]) ret++;

     printf("%d\n", ret);

 }

 int main()

 {

     while(scanf("%d", &n) ==  && n)

     {

         init();

         read();

         solve();

     }

     return ;

 }