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)

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;

typedef long long ll;
typedef long double ld;
template <typename T>
    using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template <char cmp = '<'>
    bool icmp(ll a, ll b){ if(cmp=='>') return a > b; else return a < b;}
namespace enhanced_pair{
    typedef pair<ll,ll> pll;
    ostream& operator << (ostream& o, const pll a){
        o<<a.first<<" "<<a.second;
        return o;
    }
    istream& operator >> (istream& i, pll& p){
        i>>p.first>>p.second;
        return i;
    }
    template <char cmp1 = '<', char cmp2 = '<'>
        bool pcmp(pll a, pll b){ return icmp<cmp1> (a.first,b.first) || (a.first==b.first && icmp<cmp2> (a.second,b.second));}
    template <char cmp1 = '<', char cmp2 = '<'>
        struct pcmpobj{bool operator()(pll a, pll b){return pcmp<cmp1,cmp2>(a,b);}};
} using namespace enhanced_pair;
const ll BIG = 1e18;
const ll NMAX = 2e5+5;
const ll MOD = 1e9+7;
map<queue<char>,array<int,26>> um;
string s[NMAX];
void tc(){
    ll n;
    cin>>n;
    for(ll i=0;i<n;i++){
        cin>>s[i];
        queue<char> q;
        for(ll j=1;j<s[i].size();j++)
            q.push(s[i][j]);
        um[q][s[i][0]-'a']++;
    }
    ll cnt=0;
    for(ll i=0;i<n;i++){
        int freq[26]={0};
        queue<char> q;
        for(auto it : s[i])
            q.push(it);
        for(ll j=0;j<=s[i].size();j++){
            auto& freq2=um[q];
            for(ll k=0;k<26;k++)
                cnt+=min(freq[k],freq2[k]);
            if(j<s[i].size()){
                freq[s[i][j]-'a']++;
                q.pop();
            }
        }
    }
    cout<<cnt-n;
}
int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    //ll q; cin>>q; while(q--)
        tc();

    return 0;
}