/* It is just a binary counter, counting down from an initial pattern */
/* The initial pattern is the answer, but what if they don't realize? */
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>

typedef long long ll;

// return 0 if no more ocelots 
int bell(char *critters,int N)
{
  //find lowest ocelot
  char *begin; 
  if((begin = strchr(critters,'O')) == NULL)
    return 0;
  *(begin--) = 'Z';
  while(begin >= critters)
    *(begin--) = 'O';
  return 1;
}
	 
void printstr(char *critters,int N)
{
  do{
    printf("%c",critters[N--]);
  }while(N>=0);
  printf("\n");
}
int main()
{
  ll bits = 0;
  ll mask;
  int N,i;   // ocelots are 1, zebras are 0
  int line = 0;
  int ch;
  ll count=0;
  char *critters;
  if(scanf("%d",&N)!=1)
    {
      perror("error reading N:");
      exit(1);
    }
  ++line;
  critters = malloc(sizeof(char)*(N+1));
  for(i=0;i<N;i++)
    {
      ++line;
      while(isspace(ch = getchar()));
      if(ch != 'O' && ch != 'Z')
	{
	  fprintf(stderr,"Line %d: ",line);
	  perror("error reading animal: ");
	  exit(2);
	}
      critters[N-i-1] = ch; // reverse order in the array
    }
  critters[N]=0; // null terminator


  //printstr(critters,N);
  while(bell(critters,N))
    {
      count++;
      //printstr(critters,N);
    }
  printf("%lld\n",count);
  return 0;
}