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: Accepted)

#include<bits/stdc++.h>
#define MAXN 2000005
#define INF 1000000000
#define MOD 1000000007
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
string str[MAXN];
int tot=1,n;
int trie[MAXN][26],cnt[MAXN][26];
bool ed[MAXN];
bool has[26];
void insert(string s,int rt)
{
    vector<int> nodes;
    nodes.clear(); 
    for(int i=0;i<(int)s.size();i++)
    {
        int x=s[i]-'a';
        if(trie[rt][x]==0) trie[rt][x]=++tot;
        nodes.push_back(rt);
        rt=trie[rt][x];
    }
    memset(has,false,sizeof(has));
    for(int i=(int)s.size()-1;i>=0;i--)
    {
        has[s[i]-'a']=true;
        int cur=nodes[i];
        for(int j=0;j<26;j++) if(has[j]) cnt[cur][j]++;
    }
}
int get_ans(string s,int rt)
{
    for(int i=0;i<(int)s.size()-1;i++)
    {
        int x=s[i]-'a';
        if(trie[rt][x]==0) return false;
        rt=trie[rt][x];
    }
    int x=s[s.size()-1]-'a';
    return cnt[rt][x];
}
map<string,int> mp;
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        cin>>str[i];
        reverse(str[i].begin(),str[i].end());
        insert(str[i],1);
    }
    ll ans=0;
    for(int i=0;i<n;i++) ans+=get_ans(str[i],1)-1;
    for(int i=0;i<n;i++) mp[str[i]]++;
    for(auto p:mp) ans-=1LL*p.S*(p.S-1)/2;
    printf("%lld\n",ans);
}