codrecorces B. Far 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

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

output

2

input

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

直接利用区间覆盖，暴力求解

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int a[2][400];int main(){int t,l,r,i,j;char ch[10];while(~ scanf("%d",&t)){memset(a,0,sizeof(a));for(i = 0 ; i < t ; i++){scanf("%s %d %d",ch,&l,&r);if(ch[0] == 'M'){for(j = l ; j <= r ; j++)   a[0][j]++;}else{for(j = l ; j <= r ; j++)   a[1][j]++;}}int ans = 0;for( i = 0 ; i <= 366 ; i++)ans=max(ans,2*min(a[0][i],a[1][i]));printf("%d\n",ans);}return 0; }