Far Relative’s Problem

Description

Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has n friends and each of them can come to the party in a specific range of days of the year from a_i to b_i. Of course, Famil Door wants to have as many friends celebrating together with him as possible.

Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party.

Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends that may present at the party.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 5000) — then number of Famil Door's friends.

Then follow n lines, that describe the friends. Each line starts with a capital letter 'F' for female friends and with a capital letter 'M' for male friends. Then follow two integers a_i and b_i (1 ≤ a_i ≤ b_i ≤ 366), providing that the i-th friend can come to the party from day a_i to day b_i inclusive.

Output

Print the maximum number of people that may come to Famil Door's party.

Sample Input

Input

4
M 151 307
F 343 352
F 117 145
M 24 128

Output

2

Input

6
M 128 130
F 128 131
F 131 140
F 131 141
M 131 200
M 140 200

Output

4

Hint

In the first sample, friends 3 and 4 can come on any day in range [117, 128].

In the second sample, friends with indices 3, 4, 5 and 6 can come on day 140.

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct  node
{
	char sex;
	int  begin,end;
}zu[20001];
int nv[2001];
int nan[2001];
int min(int a,int b)
{
	if(a<=b)
	return a;
	else
	return b; 
}
int main()
{
		memset(nv,0,sizeof(nv));
			memset(nan,0,sizeof(nan));
		 int n;
		 scanf("%d",&n);
		 for(int i=0;i<n;i++)
		 {
		 		getchar();
		 	scanf("%c %d%d",&zu[i].sex,&zu[i].begin,&zu[i].end);
		 }
		 for(int i=0;i<n;i++)
		 {
		    for(int j=zu[i].begin ;j<=zu[i].end ;j++)
		 		{ 
				     if(zu[i].sex%2==1)
		 	     	{
		 				nv[j]+=1;	
				     }	 	
				      else
				      {
				     	nan[j]+=1;
				       }		      	
		         }
		 }
		 int max=0,sum;	 
		 for(int i=1;i<=366;i++)
		 {
		      sum=min(nv[i],nan[i]);
		      if(sum>max)
		      {
		      	max=sum;
			  }
		 }
		 printf("%d
",max*2);
	return 0;
}