Far Relative’s Problem

 Far Relative’s Problem

 

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.

Example

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

第一眼看到这道题以为是区间覆盖问题，就上网学了区间覆盖类问题的解题方法，结果这道题还是不会做（可能我自己的思路不正确）。最后还是看大佬们的博客才弄懂其中有一段代码卡了一上午才想明白是咋回事。弄懂了才发现这道题就是简单的贪心（可能贪心都算不上），处理方法有点巧妙（还是要多刷题啊）。废话不多说，讲解奉上。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int arr[500],crr[500],brr[500];
bool cmp1(int a,int b){
	return a>b;
}
int main(){
	int t;
	int a,b;
	char c[2];
	while(scanf("%d",&t)!=EOF){
		memset(arr,0,sizeof(arr));
		memset(brr,0,sizeof(brr));
		memset(crr,0,sizeof(crr));
		while(t--){
			scanf("%s %d %d",&c,&a,&b);
			if(c[0]=='M'){
				for(int i=a;i<=b;i++) arr[i]++;/*用一个数组用来记录男朋友有时间的
				那一天的出现次数，以此来找出男朋友之间有时间最多的那一天 */ 
			}
			if(c[0]=='F'){
				for(int i=a;i<=b;i++) brr[i]++;//同上 
			}
		}
		for(int i=0;i<367;i++){
				crr[i]=min(arr[i],brr[i]);/*当时就是这段代码没想明白
				这里我着重讲解，找出男朋友女朋友有共同时间最少的那一天
				并把其储存到数组中。为什么要找最少的那一天？因为题目中
				说了来的男朋友女朋友要一样多，就要考虑把多余的一部分人
				删去。*/ 
			}
		sort(crr,crr+500,cmp1);
		printf("%d\n",crr[0]*2);
	}
	return 0;
}