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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define REP(i,m,n) for(int i=(m);i<(n);i++)
#define rep(i,n) REP(i,0,n)
#define pb push_back
#define all(a) a.begin(),a.end()
#define rall(c) (c).rbegin(),(c).rend()
#define mp make_pair
#define endl '\n'
#define vec vector<ll>
#define mat vector<vector<ll> >
#define fi first
#define se second
typedef long long ll;
typedef pair<ll,ll> pll;
typedef long double ld;
const ll inf=1e9+7;
const ll mod=1e4+7;

int MAX=200010;
vector<vector<int> >trie(MAX,vector<int>(26));
vector<int>leaf(MAX);

int sz=1;
void add_string(string s){
    int v=0;
    for(auto c:s){
        if(!trie[v][c-'a']){
            trie[v][c-'a']=sz;
            sz++;
        }
        v=trie[v][c-'a'];
    }
    leaf[v]=1;
}

signed main(){
    ll n;cin>>n;
    vector<string>a(n);
    rep(i,n){
        cin>>a[i];
        reverse(all(a[i]));
        add_string(a[i]);
    }
    ll ans=0;
    rep(i,n){
        string s;
        s=a[i];
        vector<ll>cnt(26);
        for(auto c:s){
            cnt[c-'a']++;
        }
        ll v=0;
        rep(j,s.size()){
            rep(x,26){
                if(cnt[x]&&trie[v][x]){
                    ans+=leaf[trie[v][x]];
                }
            }
            cnt[s[j]-'a']--;
            v=trie[v][s[j]-'a'];
        }
        ans--;
    }
    cout<<ans<<endl;
}