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)

#include <bits/stdc++.h>
#define mk make_pair
#define fs first
#define sc second
using namespace std;
typedef long long ll;
typedef long double ld;
// please, read the question correctly (do you need set or multiset)???
const int N=200010; //check the limits, dummy
int a[N];
int n, m, cnt[44];
char s[1000010];
// struct hash_pair { 
//     template <class T1, class T2> 
//     size_t operator()(const pair<T1, T2>& p) const
//     { 
//         auto hash1 = hash<T1>{}(p.first); 
//         auto hash2 = hash<T2>{}(p.second); 
//         return hash1*1000000000+hash2; 
//     } 
// };

unordered_map<int, int> mp;
// int mp[33][1]
string c[N];
int main(){
	scanf("%d",&n);
	ll ans=0, mod=1e9+9, mod1=1e9+9;
	for(int i=0; i<n; ++i){
		scanf("%s",s);
		// for(int j=0; j<100000; ++j){
		// 	s[j]='a';
		// }
		// s[100000]='\0';
		int sz=strlen(s);
		c[i]=s;
		ll hsh=0, hsh1=0;
		for(int j=sz-1; j>=0; --j){
			// cout<<s[j]-'a'<<endl;
			hsh*=29;
			hsh+=s[j]-'a'+1;
			hsh%=mod;
					// while(hsh>=mod)hsh-=mod;
			// hsh1*=29;
			// hsh1+=s[j]-'a';
			// hsh1%=mod1;
		}
		// cout<<endl;
		++mp[hsh];
		// cout<<hsh<<" "<<hsh1<<endl;
	}
	for(int i=0; i<n; ++i){
		// cout<<c[i]<<endl;
		int sz=c[i].size();
		for(int j=0; j<26; ++j)
			cnt[j]=0;
		for(int j=0; j<sz; ++j){
			++cnt[c[i][j]-'a'];
		}
		ll hsh=0, hsh1=0, tmp1, tmp2;
		for(int j=sz-1; j>=0; --j){
			for(int k=0; k<26; ++k){
				if(cnt[k]){
					// cout<<k<<"&"<<endl;
					tmp1=hsh*29;
					tmp1+=k+1;
					// while(tmp1>=mod)tmp1-=mod;
					tmp1%=mod;
					// tmp2=hsh1*29;
					// tmp2+=k;
					// tmp2%=mod1;
					ans+=mp[tmp1];
					// cout<<"#"<<tmp1<<" "<<tmp2<<endl;cout<<mp[mk(tmp1, tmp2)]<<endl;
				}
			}
			// cout<<c[i][j]-'a'<<"(((((((((("<<endl;
			hsh*=29;
			hsh+=c[i][j]-'a'+1;
			hsh%=mod;
					// while(hsh>=mod)hsh-=mod;
			// hsh1*=29;
			// hsh1+=c[i][j]-'a';
			// hsh1%=mod1;
			--cnt[c[i][j]-'a'];
		}
	}
	ans-=n;
	cout<<ans<<endl;
}