Problem:
Score : 700 points

Problem Statement

Limak can repeatedly remove one of the first two characters of a string,
for example abcxyx \rightarrow acxyx \rightarrow cxyx \rightarrow cyx.

You are given N different strings S_1, S_2, \ldots, S_N.
Among N \cdot (N-1) / 2 pairs (S_i, S_j),
in how many pairs could Limak obtain one string from the other?

Constraints

2 \leq N \leq 200\,000

S_i consists of lowercase English letters a-z.

S_i \neq S_j

1 \leq |S_i|

|S_1| + |S_2| + \ldots + |S_N| \leq 10^6

Input

Input is given from Standard Input in the following format.

N
S_1
S_2
\vdots
S_N

Output

Print the number of unordered pairs (S_i, S_j)
where i \neq j and Limak can obtain one string from the other.

Sample Input 1

3
abcxyx
cyx
abc

Sample Output 1

1

The only good pair is (abcxyx, cyx).

Sample Input 2

6
b
a
abc
c
d
ab

Sample Output 2

5

There are five good pairs: (b, abc), (a, abc), (abc, c), (b, ab), (a, ab).

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

        #include <bits/stdc++.h>
        using namespace std;
   #define int long long
    const int MAX =200002;
const int MOD =1000000007;
int mp[('z'-'a')+1][1000001];
     signed main(){
int  n;

cin>>n;
string s[n],ss;
pair<string,char> p[n];
for(int i=0;i<n;i++){
    cin>>s[i];
   reverse(s[i].begin(),s[i].end());
   char c=s[i][s[i].size()-1];
   s[i].pop_back();
   p[i]={s[i],c};
}
sort(p,p+n);
int ans=0;
for(int y=0;y<=('z'-'a');y++)mp[y][0]=0;
for(int i=0;i<n;i++){
    int k=0;
    s[i]=p[i].first;
    s[i].push_back(p[i].second);
    bool t[('z'-'a')+1]={};
    while(i!=0&&k<s[i-1].size()&&k<s[i].size()&&s[i-1][k]==s[i][k])k++;
    for(int j=k+1;j<s[i].size();j++)for(int y=0;y<=('z'-'a');y++)mp[y][j]=0;
    for(int j=s[i].size()-1;j>=0;j--){
        t[(s[i][j]-'a')]=true;
       for(int jj=0;jj<=('z'-'a');jj++)if(t[jj]) ans+=mp[jj][j];
    }
    mp[(s[i][s[i].size()-1]-'a')][s[i].size()-1]++;
}
cout<<ans<<endl;
//for(int i=0;i<n;i++)cerr<<s[i]<<endl;

     }