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)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
//#include<boost/multiprecision/cpp_int.hpp>
//#include<boost/multiprecision/cpp_dec_float.hpp>
//namespace mp=boost::multiprecision;
//#define mulint mp::cpp_int
//#define mulfloat mp::cpp_dec_float_100
using namespace std;
struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}} __init;
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
constexpr int MOD=1000000007;
//constexpr int MOD=998244353;
#define INF (1<<30)
#define LINF (lint)(1LL<<56)
#define endl "\n"
#define rep(i,n) for(lint (i)=0;(i)<(n);(i)++)
#define reprev(i,n) for(lint (i)=(n-1);(i)>=0;(i)--)
#define Flag(x) (1<<(x))
#define Flagcount(x) __builtin_popcount(x)
#define pint pair<int,int>
#define pdouble pair<double,double>
#define plint pair<lint,lint>
#define fi first
#define se second
typedef long long lint;
int dx[8]={1,1,0,-1,-1,-1,0,1};
int dy[8]={0,1,1,1,0,-1,-1,-1};
const int MAX_N=2e5+5;

int main(void){
    int N;
    cin >> N;
    string S[N];
    rep(i,N) cin >> S[i];
    rep(i,N) reverse(S[i].begin(),S[i].end());
    lint place[N][26]={};
    rep(i,N) rep(j,26) place[i][j]=-INF;
    sort(S,S+N);
    rep(i,N) rep(j,S[i].length()) place[i][S[i][j]-'a']=max(place[i][S[i][j]-'a'],j);
    lint ans=0;
    rep(i,N){
        auto itr=lower_bound(S,S+N,S[i].substr(0,S[i].length()-1)+(char)(S[i][S[i].length()-1]+1));
        ans+=itr-(S+i)-1;
    }
    rep(i,N){
        int alf[26]={};
        for(int j=S[i].length();j>=0;j--){
            rep(k,26) if(alf[k]){
                auto itr=binary_search(S,S+N,S[i].substr(0,j)+(char)('a'+k));
                if(itr==true) ans++;
            }
            alf[S[i][j]-'a']++;
        }
    }
    cout << ans << endl;
}