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: Time Limit Exceeded)

//@formatter:off
#include<bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < int(n); ++i)
#define rrep(i,n) for (int i = int(n)-1; i >= 0; i--)
#define rep2(i,s,n) for (int i = int(s); i < int(n); ++i)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define pb push_back
#define eb emplace_back
#define vi vector<int>
#define vvi vector<vector<int>>
#define vl vector<ll>
#define vvl vector<vector<ll>>
#define vd vector<double>
#define vvd vector<vector<double>>
#define vs vector<string>
#define vc vector<char>
#define vvc vector<vector<char>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define vp vector<P>
#define vvp vector<vector<P>>
using namespace std;
using ll = long long;
using P = pair<int,int>;
using LP = pair<ll,ll>;
template<class S,class T> istream& operator>>(istream &is,pair<S,T> &p) { return is >> p.first >> p.second; }
template<class S,class T> ostream& operator<<(ostream &os,const pair<S,T> &p) { return os<<'{'<<p.first<<","<<p.second<<'}'; }
template<class T> istream& operator>>(istream &is,vector<T> &v) { for(T &t:v){is>>t;} return is; }
template<class T> ostream& operator<<(ostream &os,const vector<T> &v) { os<<'[';rep(i,v.size())os<<v[i]<<(i==int(v.size()-1)?']':','); return os; }
void Yes(bool b) { cout << (b ? "Yes" : "No") << '\n'; }
void YES(bool b) { cout << (b ? "YES" : "NO") << '\n'; }
template<class T> bool chmin(T& a,T b) {if(a > b){a = b; return true;} return false;}
template<class T> bool chmax(T& a,T b) {if(a < b){a = b; return true;} return false;}
const int inf = 1001001001;
const ll linf = 1001001001001001001;
//@formatter:on

using T = pair<int,string>;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);
    int n;
    cin >> n;
    vs v(n);
    cin >> v;
    ll ans = 0;
    map<string,int> mp;
    rep(i,n) mp[v[i]]++;
    for(auto [f,s] : mp) ans += (ll)s*(s-1)/2;
    rep(i,n) reverse(all(v[i]));
    sort(all(v), [](string a, string b){ return a.size() < b.size(); });
    map<T,int> cnt;
    for(string s : v) {
        string t = s;
        vb ex(26);
        while(t.size()) {
            char c = t.back();
            t.pop_back();
            ex[c-'a'] = true;
            if(t.size() == s.size()-1) continue;
            rep(i,26) if(ex[i]) {
                if(cnt.count({i,t})) ans += cnt[{i,t}];
            }
        }
        string now = s;
        now.pop_back();
        char c = s.back();
        cnt[{c-'a',now}]++;
    }
    cout << ans << endl;
}