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 ti 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, ..., tn, 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.


题意:给n个字符串,仅由s和h组成,不改变每个字符串的内部顺序,但可以改变字符串的出现顺序.问最多能出现多少个sh子序列???

思路:假设现在对比2个字符串s1,s2  .   那么如果  s1.cnts*s2.cnth>s2.cnts*s1.cnth 那么说明s1在s2前面是最好的,移项可知,串中s:h比例最大的放前面对答案的贡献值最大.

#include <bits/stdc++.h>
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define bug cout << "bug" << endl
#pragma comment(linker, "/STACK:102400000,102400000")


using namespace std;
typedef long long ll;

const int MAX_N=1e5+5;

struct node{
    string s;
    int cnts,cnth;
    double rate;
    node(){
        cnts=0,cnth=0,rate=0;
    }
}p[MAX_N];

bool cmp(const node a,const node b){
    return a.rate>b.rate;
}

int main(void){
    int n;
    cin >> n;
    ll toth=0;
    for(int i=1;i<=n;i++){
        cin >> p[i].s;
        for(int j=0;j<(int)p[i].s.size();++j){
            if(p[i].s[j]=='s'){
                p[i].cnts++;
            }
            else toth++,p[i].cnth++;
        }
        if(p[i].cnth==0)    p[i].rate=INF;
        else p[i].rate=(double)p[i].cnts/p[i].cnth;
    }
    sort(p+1,p+1+n,cmp);
    ll ans=0;
    for(int i=1;i<=n;i++){
        for(int j=0;j<(int)p[i].s.size();++j){
            if(p[i].s[j]=='s')  ans+=toth;
            else    toth--;
        }
    }
    cout << ans << endl;
}