#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <map>
#include <cmath>
#include <vector>
#include <fstream>
using namespace std;

/*Resistors connected to each point. First = the point on the other end of the resistor,
second = the resistance value.*/
vector<map<int, double> > resistors(26);

//Number of neighbors each point has.
int numNeighbors[26];

/* Inserts a resistor connecting points "begin" and "end" with resistance "value". If
a resistor already exists there, inserts the new resistor in parallel. */
void putRes(int begin, int end, double value)
{
     if (resistors[begin][end] > 0) //A resistor already exists, insert in parallel
     {
        resistors[begin][end] = 1.0/(1.0/resistors[begin][end] + 1.0/value);
     }
     else //This is the only resistor between the 2 points
     {
         resistors[begin][end] = value;
         numNeighbors[begin]++;
     }
}

int main()
{
    ifstream infile("circuits.judge");
    ofstream outfile("circuits.out");
    
    int circuitNum = 0;
    
    while(true)
    {
        circuitNum++;
               
        //Reset the data structures
        for (int i = 0; i < 26; i++)
        {
            numNeighbors[i] = 0;
            resistors[i].clear();
        }
    
        int numResistors;
        infile >> numResistors;
        if (numResistors == 0)
           break;
        
        for (int i = 0; i < numResistors; i++)
        {
            char ch1, ch2;
            double value;
            infile >> ch1 >> ch2 >> value;
            
            int point1 = ch1 - 'A'; //Convert characters to integers for easy storage
            int point2 = ch2 - 'A';
            putRes(point1, point2, value); //Keep track of the resistor in both directions
            putRes(point2, point1, value);
        }
    
        bool reduced = false; //Has anything been reduced on this loop iteration?
        do
        {
         reduced = false;
         for (int i = 1; i < 25; i++) //B through Y - don't want to eliminate A or Z
         {
             //The point is only connected to 2 resistors, so they can be combined in series.
             if (numNeighbors[i] == 2) 
             {   
                reduced = true;
                int begin = (resistors[i].begin())->first; //one of the point's neighbors
                int end = (resistors[i].rbegin())->first; //the other neighbor

                //Add the new, combined resistor between the 2 neighbors.
                double newRes = resistors[begin][i] + resistors[i][end];
                putRes(begin, end, newRes);
                putRes(end, begin, newRes);
                
                //This point has been reduced away and no longer exists
                resistors[i].clear();
                numNeighbors[i] = 0;
                
                //Get rid of this point on its neighbors' neighbor lists.
                resistors[begin].erase(i);
                resistors[end].erase(i);
                numNeighbors[begin]--;
                numNeighbors[end]--;

             }
         }
        } while (reduced);
            
        /* Count the number of resistors left. The count will be 2 * the actual number
        because each resistor was stored twice.*/
        int total = 0; 
        for (int i = 0; i < 26;  i++)
        {
            total += numNeighbors[i];
        }
        
        if (total > 2 || //we weren't able to fully reduce the circuit
           resistors[0].find(25) == resistors[0].end()) //points A and Z are not connected
           outfile << "Circuit " << circuitNum << ": -1.000" << endl;
        else
        {
            double finalRes = resistors[0][25];
            outfile << setprecision(3) << fixed;
            outfile << "Circuit " << circuitNum << ": " << finalRes << endl;
        }
    }
    return 0;
}