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)

/*
    Mbak Sana Dong
    Mbak Chaeyoung Dong
*/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll mod = 1e9+9;
const ll p = 31;
string arr[200200];
vector<pair<int,int> > v;
unordered_map<ll,ll> mp;
int cnt[33];
int main()
{
    int n,i,j,k,x,y,ans,curr,temp,m;
    int tc;
    cin >> n;
    for(i=1;i<=n;i++) {
        cin >> arr[i];
        reverse(arr[i].begin(), arr[i].end());
    }
    ll chaeng = 0;
    vector<ll> to;
    for(k=1;k<=n;k++) {
        int sz = arr[k].size();
        int idx = k;
        for(i=1;i<=26;i++) cnt[i] = 0;
        for(auto x : arr[idx]) cnt[x - 'a' + 1]++;
        ll now = 0, pp = p;
        for(i=1;i<=26;i++) {
            if(cnt[i] > 0) mp[i%mod]++;
        }
        for(i=0;i<sz-1;i++) {
            now += ((arr[idx][i]-'a'+1)*pp%mod);
            now %= mod;
            pp *= p;
            pp %= mod;
            cnt[arr[idx][i]-'a'+1]--;
            for(j=1;j<=26;j++) {
                if(cnt[j] > 0) mp[(now+((1LL*j*pp)%mod))%mod]++;
            }
        }
        now += ((arr[idx][sz-1]-'a'+1)*pp % mod);
        now %= mod;
        if(sz==1) now/=pp;
        to.push_back(now);
    }
    for(auto x : to) {
        mp[x]--;
        chaeng += mp[x];
        mp[x]++;
    }
    cout << chaeng << endl;
    return 0;
}