Codeforce922D (Robot Vacuum Cleaner)

D. Robot Vacuum Cleaner

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Pushok the dog has been chasing Imp for a few hours already.

Fortunately, Imp knows that Pushok is afraid of a robot vacuum cleaner.

While moving, the robot generates a string t consisting of letters 's' and 'h', that produces a lot of noise. We define noise of string t as the number of occurrences of string "sh" as a subsequence in it, in other words, the number of such pairs (i, j), that i < j and and .

The robot is off at the moment. Imp knows that it has a sequence of strings t_i in its memory, and he can arbitrary change their order. When the robot is started, it generates the string t as a concatenation of these strings in the given order. The noise of the resulting string equals the noise of this concatenation.

Help Imp to find the maximum noise he can achieve by changing the order of the strings.

Input

The first line contains a single integer n (1 ≤ n ≤ 10⁵) — the number of strings in robot's memory.

Next n lines contain the strings t₁, t₂, ..., t_n, one per line. It is guaranteed that the strings are non-empty, contain only English letters 's' and 'h' and their total length does not exceed 10⁵.

Output

Print a single integer — the maxumum possible noise Imp can achieve by changing the order of the strings.

Examples

Input

Copy

4
ssh
hs
s
hhhs

Output

18

Input

Copy

2
h
s

Output

1

Note

The optimal concatenation in the first sample is ssshhshhhs.

分析：看代码吧=_=，懒得写了。

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
/*使h前面的s尽可能多,字符串一个一个累加 
*/
struct Node{
    string sub;
    long long s,h;
    long long cnt;
}a[200000];

int cmp(Node A,Node B)//对于不同的字符串,按照贡献值大小排序 
{
    if(A.s*B.h>A.h*B.s) return 1;
    return 0;
}

int main()
{
    int N;
    scanf("%d",&N);
    long long ans=0;
    for(int i=0;i<N;i++)
    {
        cin>>a[i].sub;
        a[i].s=a[i].h=a[i].cnt=0;
        for(int j=0;j<(a[i].sub).size();j++)//计算每个字符串自己的贡献值 
        {
            if(a[i].sub[j]=='h') {a[i].h++;a[i].cnt+=a[i].s;}
            else a[i].s++;
        }
        ans+=a[i].cnt;
    }
    
    sort(a,a+N,cmp);//排序 
    long long sum=a[0].s;
    for(int i=1;i<N;i++)
    {
        ans+=sum*a[i].h;
        sum+=a[i].s;
    }
    
    printf("%lld
",ans);
    return 0;
}