/*
 * @(#) msu96_f.c - Problem 'F' ("A game") solution
 * of the ACM Programming Contest at the MSU in 1996.
 * (c) 1996 Ivan Maidanski <ivmai@chat.ru> http://ivmai.chat.ru
 * Freeware program source. All rights reserved.
 **
 * Language: ANSI C
 * Tested with: Borland C++ v3.1
 * Last modified: 1996-12-02 14:00:00 GMT+03:00
 */

/* Input data file: f.dat */

#include <stdio.h>
#include <ctype.h>
#include <string.h>

char *lex[]=
{
 "null","cheat","trade",
 "if","then","else",
 "=","#","and","or",
 "mylast1","yourlast1","mylast2","yourlast2",
 0
};

char prog[10][256];

int readprog(void)
{
 int i,n;
 char c;
 FILE *f=fopen("f.dat","rt");
 for (n=0;n<10;n++)
 {
  for (i=0;i<255;i++)
  {
   do
    if ((c=getc(f))==EOF)
     c='#';
    while (c<=' ');
   if (c=='.' || c=='#' && !i)
    break;
   prog[n][i]=isupper(c) ? tolower(c) : c;
  }
  if (c=='#' && !i)
   break;
  prog[n][i]='\0';
 }
 fclose(f);
 return n;
}

int getlexem(char **s)
{
 int k;
 if (!**s)
  return -1;
 for (k=0;lex[k];k++)
  if (!strncmp(*s,lex[k],strlen(lex[k])))
   break;
 (*s)++;
 if (!lex[k])
  return -1;
 (*s)+=strlen(lex[k])-1;
 return k;
}

int interpret(char *s, int ml1, int ml2, int yl1, int yl2)
{
 int k,mem,cond;
 while (*s)
 {
  if ((k=getlexem(&s))<3)
   return k-1;
  cond=0;
  do
  {
   switch (getlexem(&s))
   {
    case 10:
     mem=ml1;
     break;
    case 11:
     mem=yl1;
     break;
    case 12:
     mem=ml2;
     break;
    case 13:
     mem=yl2;
     break;
    default:
     return -2;
   }
   k=getlexem(&s);
   mem-=getlexem(&s)-1;
   if (!cond)
    cond=(k==6 && !mem || k!=6 && mem);
   k=getlexem(&s);
   if (cond<0)
   {
    if (k==6)
     cond++;
   }
    else
    {
     if (k==8)
      cond--;
    }
  } while (k!=4);
  if (cond<1)
  {
   cond=0;
   while (*s && cond>=0)
    if ((k=getlexem(&s))==3)
     cond++;
     else if (k==5)
      cond--;
  }
 }
 return -2;
}

void game(int i, int j)
{
 int k;
 int my,your;
 int ml1=-1,yl1=-1,ml2=-1,yl2=-1;
 int mr=0,yr=0;
 for (k=0;k<10;k++)
 {
  my=interpret(prog[i],ml1,ml2,yl1,yl2);
  your=interpret(prog[j],yl1,yl2,ml1,ml2);
  if (my<0 || your<0)
  {
   printf("error\n");
   break;
  }
  if (my)
   if (your)
    mr++,yr++;
    else mr-=2,yr+=2;
   else if (your)
    mr+=2,yr-=2;
    else mr--,yr--;
  ml2=ml1;
  yl2=yl1;
  ml1=my;
  yl1=your;
 }
 printf("%d:%d\n",mr,yr);
}

int main()
{
 int i,j;
 int n=readprog();
 for (i=0;i<n;i++)
  for (j=i+1;j<n;j++)
   game(i,j);
 return 0;
}