//O(M * 2^M), here M = 2 * N.
#include <algorithm> // if you have problems with this C++ code,
#include <cmath> // consult your programming text books first...
#include <cstdio>
#include <cstring>
using namespace std;
int N, target; // but it is OK for competitive programming
double dist[20][20], memo[1 << 16]; // 1 << 16 = 2^16, note that max N = 8
double matching(int bitmask) { // DP state = bitmask
// we initialize ‘memo’ with -1 in the main function
	if (memo[bitmask] > -0.5) // this state has been computed before
		return memo[bitmask]; // simply lookup the memo table
	if (bitmask == target) // all students are already matched
		return memo[bitmask] = 0; // the cost is 0
	double ans = 2000000000.0; // initialize with a large value
	int p1, p2;
	for (p1 = 0; p1 < 2 * N; p1++)
		if (!(bitmask & (1 << p1)))
			break; // find the first bit that is off
	for (p2 = p1 + 1; p2 < 2 * N; p2++) // then, try to match p1
		if (!(bitmask & (1 << p2))) // with another bit p2 that is also off
			ans = min(ans, // pick the minimum
					dist[p1][p2] + matching(bitmask | (1 << p1) | (1 << p2)));
	return memo[bitmask] = ans; // store result in a memo table and return
}
int main() {
	int i, j, caseNo = 1, x[20], y[20];
	while (scanf("%d", &N), N) { // yes, we can do this :)
		for (i = 0; i < 2 * N; i++)
			scanf("%*s %d %d", &x[i], &y[i]); // ’%*s’ skips names
		for (i = 0; i < 2 * N - 1; i++) // build pairwise distance table
			for (j = i + 1; j < 2 * N; j++) // have you used ‘hypot’ before?
				dist[i][j] = dist[j][i] = hypot(x[i] - x[j], y[i] - y[j]);
				// use DP to solve min weighted perfect matching on small general graph
		for (i = 0; i < (1 << 16); i++) memo[i] = -1.0; // set -1 to all cells
		target = (1 << (2 * N)) - 1;
		printf("Case %d: %.2lf\n", caseNo++, matching(0));
	}
}