codeforces 629BFar Relative’s Problem

B. Far Relative’s Problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ai to bi. 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 ai and bi (1 ≤ ai ≤ bi ≤ 366), providing that the i-th friend can come to the party from day ai to day biinclusive.

Output

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

Examples

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

Note

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.

题意：家里接待客人但是每次必须接待一对夫妻，即必须是一男一女，现在给出一年内一些人到作者家里做客的区间，问作者家一天最多可能会接待多少人

题解：显然此题是要求最多次重复的区间，并且在这个重复的区间内，男性的人数等于女性的人数，

#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 5100
#define mod 10007
#define dian 1.000000011
#define INF 0x3f3f3f
using namespace std;
int main()
{
	int n,j,i,t;
	char op[3];
	while(scanf("%d",&n)!=EOF)
	{
		int m[MAX],f[MAX];
		int beg,end;
		memset(m,0,sizeof(m));
		memset(f,0,sizeof(f));
		for(i=1;i<=n;i++)
		{
			scanf("%s%d%d",op,&beg,&end);
			if(op[0]=='M')
				for(j=beg;j<=end;j++)
				    m[j]++;
			else
			    for(j=beg;j<=end;j++)
			        f[j]++;
		}
		int sum=-INF;
		int F,M;
		F=M=0;
		for(i=1;i<=366;i++)
			sum=max(sum,min(m[i],f[i]));
		printf("%d
",sum*2);
	}
	return 0;
}