// Solution to "square count" by Bob Roos
//
// Solution just uses formulas I came up with for number of squares
// contained within a rectangular room and number of squares that
// overlap a pair of rooms.

import java.util.*;
import java.io.*;

public class BSquareCount {
 public static BufferedReader in;
 public static StringTokenizer tok;
 public static int n;
 public static int[] x1,y1,x2,y2;
 public static int count;
 public static int caseNum;

 public static void main(String[] args) throws IOException {
   in = new BufferedReader(new InputStreamReader(System.in));
   String line;
   caseNum = 0;
   while (!(line = in.readLine().trim()).equals("0")) {
     n = Integer.parseInt(line);
     caseNum++;
     x1 = new int[n]; y1 = new int[n]; x2 = new int[n]; y2 = new int[n];
     for (int i = 0; i < n; i++) {
       line = in.readLine().trim();
       tok = new StringTokenizer(line);
       x1[i] = Integer.parseInt(tok.nextToken());
       y1[i] = Integer.parseInt(tok.nextToken());
       x2[i] = Integer.parseInt(tok.nextToken());
       y2[i] = Integer.parseInt(tok.nextToken());
       // make (x1,y1) the lower left corner:
       if (x1[i] > x2[i]) {
         int temp = x1[i]; x1[i] = x2[i]; x2[i] = temp;
         temp = y1[i]; y1[i] = y2[i]; y2[i] = temp;
       }
       if (y1[i] > y2[i]) {
         int temp = y1[i]; y1[i] = y2[i]; y2[i] = temp;
       }
     }
     System.out.println("Case " + caseNum + ": " + sqCount());
   }
 }

 public static int sqCount() {
   count = 0;
   // first, count only squares limited to a single room:
   for (int i = 0; i < n; i++) {
     int width = x2[i]-x1[i];
     int height = y2[i]-y1[i];
     // formula assumes width >= height, so...
     if (width < height) {
       int temp = width; width = height; height = temp;
     }

     // Following kludge prevents overflow when height and width are
     // large:

//ORIGINAL:     int k = height*(height+1)*(2*height+1)/6;
     int k = 1;
     if (height % 2 == 0)
       k *= (height/2) * (height+1);
     else
       k *= (height+1)/2 * height;
     if (k % 3 == 0) {
       k /= 3;
       k *= (2*height+1);
     }
     else
       k *= (2*height+1)/3;
// ORIGINAL:     k += (width-height)*(height)*(height+1)/2;
     if (height%2 == 0)
        k+= (height/2)*(width-height)*(height+1);
     else
        k+= ((height+1)/2)*(width-height)*height;
     count += k;
// System.out.println("DEBUG: room["+i+"] has " + k + " squares");
   }

   // Now for pairs of rooms. Since n <= 1000,
   // I'll just check all pairs of rooms for shared opening.
   for (int i = 0; i < n-1; i++) {
     for (int j = i+1; j < n; j++) {
       // First look at horizontal walls:
        int lower = i, upper = j;
        if (y1[i] == y2[j]) {
          lower = j; upper = i;
        }
        int xmin,xmax;
        if (y2[lower] == y1[upper]) {
          // Find overlap, if any:
          xmin = Math.max(x1[lower],x1[upper]);
          xmax = Math.min(x2[lower],x2[upper]);
          int m = xmax - xmin - 2;
          for (int k = 2; k <= m; k++) {
            if (k > y2[upper] - y1[lower])
              break;
// System.out.println("DEBUG: Rooms " + i + ", "+j+": "+
// ((m-k+1) * Math.min(Math.min(y2[upper]-y1[upper],y2[lower]-y1[lower]),k-1))
// +" "+k+"x"+k+" squares");
            count += (m-k+1) *
              Math.min(Math.min(Math.min(y2[upper]-y1[upper],
                 y2[lower]-y1[lower]),k-1),y2[upper]-y1[lower]-k+1);

          }

        }

       // Vertical walls:
        int left = i, right = j;
        if (x1[i] == x2[j]) {
          left = j; right = i;
        }
        int ymin,ymax;
        if (x2[left] == x1[right]) {
          // Find overlap, if any:
          ymin = Math.max(y1[left],y1[right]);
          ymax = Math.min(y2[left],y2[right]);
          int m = ymax - ymin - 2;
          for (int k = 2; k <= m; k++) {
            if (k > x2[right] - x1[left])
              break;
// System.out.println("Rooms " + i + ", "+j+": "+
// ((m-k+1) * Math.min(Math.min(x2[right]-x1[right],x2[left]-x1[left]),k-1))
// +" "+k+"x"+k+" squares");
            count += (m-k+1) *
                 Math.min(Math.min(Math.min(x2[right]-x1[right],
                   x2[left]-x1[left]),k-1),x2[right] - x1[left]-k+1);
          }
        }
      }
   }

   return count;
 }
}