//To print the multiplication table of a number upto an upper bound
//@author-Karan Narain(10111017)

#include <stdio.h>

int main()
{
   // d1 and d2 for dates m1 and m2 for months y1 and y2 for years and day for storing the value of  day which second date belongs  and days are intermediateresult 
   // diff is for storing the number of days between these days and result day for finding which day was date1 given date2 
   int d1,m1,y1,d2,m2,y2,day,resultday;
   long days1,days2,diff; 

   printf("enter the dates\n ");

   scanf("%d %d %d %d %d %d",&d1,&m1,&y1,&d2,&m2,&y2);

    
   
   printf("enter the day of second date\n");
   scanf("%d",&day);
   //intialize values
   days1=0;
   days2=0;
    
   //caluculating total number of days for the given dates with respect to 0/0/0 date and subtract those two values to get the difference

   // caluculating extra days in the current year leaving previous days
   switch(m1)
   {

     case 1: days1=d1; 
             break;
     
     case 2: days1=31+d1;
             break;

     case 3: days1=31+28+d1;
             break;
     
     case 4: days1=31+28+31+d1;
              break; 
        
     case 5: days1=31+28+31+30+d1;
             break;
        
     case 6: days1=31+28+31+30+31+d1;
             break;

     case 7: days1=31+28+31+30+31+30+d1;
             break;
   
     case 8: days1=31+28+31+30+31+30+31+d1;
             break;

     case 9: days1=31+28+31+30+31+30+31+31+d1;
             break;

     case 10:days1=31+28+31+30+31+30+31+31+30+d1; 
             break;

     case 11: days1=31+28+31+30+31+30+31+31+30+31+d1;
              break;

     case 12: days1=31+28+31+30+31+30+31+31+30+31+30+d1;
              break;

   }
 
  // adding one more day if y1 is leap year and month is greater than 2
  if(y1/4==0 && m1>2)
  days1=days1+1;


   // total days for date1

  days1=days1+(365*y1)+(y1/4); // y1/4 is adding one extra day for every leap year starting from 0th year
 

  // do the same thing for date2



   switch(m2)
   {

     case 1: days2=d2; 
             break;
     
     case 2: days2=31+d2;
             break;

     case 3: days2=31+28+d2;
             break;
     
     case 4: days2=31+28+31+d2;
              break; 
        
     case 5: days2=31+28+31+30+d2;
             break;
        
     case 6: days2=31+28+31+30+31+d2;
             break;

     case 7: days2=31+28+31+30+31+30+d2;
             break;
   
     case 8: days2=31+28+31+30+31+30+31+d2;
             break;

     case 9: days2=31+28+31+30+31+30+31+31+d2;
             break;

     case 10:days2=31+28+31+30+31+30+31+31+30+d2; 
             break;

     case 11: days2=31+28+31+30+31+30+31+31+30+31+d2;
              break;

     case 12: days2=31+28+31+30+31+30+31+31+30+31+30+d2;
              break;

   }
 
  // adding one more day if y2 is leap year and month is greater than 2
  if(y2/4==0 && m2>2)
  days2=days2+1;


   // total days for date1

  days2=days2+(365*y2)+(y2/4); // y2/4 is adding one extra day for every leap year starting from 0th year

    
  diff=days2-days1;
 
  // making diff as positive value
  if(diff<0)
  diff= -diff; 


  // now caluculate the day
 
 // make day between 0 to 7 for easy use of modulus and modify after


  resultday= (day+(diff%7))%7+1;    // diff%7 is for finding how many days are completed over multiple of 7 and second modulus is for wrapping the value to 0 to 6
                                     // and adjust it to 1 to 7 by adding 1


  printf("number of days are : %d",diff);

  switch(resultday)
  {

    case 1: printf("sunday");
            break;
 
    case 2: printf("monday");
            break;
 
    case 3: printf("tuesday");
            break;

    case 4: printf("wednesday");
            break;

    case 5: printf("thursday");
            break;

    case 6: printf("friday");
            break;

    case 7: printf("saturday");
            break;


  }


}