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>
#define ll long long
#define ld long double
#define ull unsigned ll
#define pint pair<int,int>
#define mk(x,y) make_pair(x,y)
#define fir first
#define sec second
#define Rep(x,y,z) for(int x=y;x<=z;++x)
#define Red(x,y,z) for(int x=y;x>=z;--x)
using namespace std;
const int MAXN=2e5+5,MAXM=1e6+5,base=13131;
char buf[1<<12],*p1=buf,*p2=buf,nc;int ny;
//inline char gc() {return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<12,stdin),p1==p2)?EOF:*p1++;}
inline char gc(){return getchar();}
inline int read(){
	int x=0;ny=1;while(nc=gc(),(nc<48||nc>57)&&nc!=EOF)if(nc==45)ny=-1;if(nc<0)return nc;
	x=nc-48;while(nc=gc(),47<nc&&nc<58&&nc!=EOF)x=(x<<3)+(x<<1)+(nc^48);return x*ny;
}
int n,vis[26];ull Fpow[MAXM];
vector<ull>Has[MAXN];
string S[MAXN];int len[MAXN];
unordered_map<ull,int>cnt[26];
int main(){
//	freopen("std.in","r",stdin);
//	freopen("std.out","w",stdout);
	n=read();
	Fpow[0]=1;Rep(i,1,MAXM-1)Fpow[i]=Fpow[i-1]*base;
	Rep(i,1,n){
		cin>>S[i];len[i]=S[i].size();
		Has[i].resize(len[i]+1),Has[i][len[i]]=0;
		for(int j=len[i]-1;j>=0;j--)Has[i][j]=Has[i][j+1]+Fpow[len[i]-j-1]*S[i][j];
		cnt[S[i][0]-'a'][Has[i][1]]++;
	}ll ans=0;
	Rep(i,1,n){
		Rep(c,0,25)vis[c]=0;
		cnt[S[i][0]-'a'][Has[i][1]]--;
		for(int j=0;j<=len[i];j++){
			Rep(c,0,25)if(vis[c])ans+=cnt[c][Has[i][j]];
			vis[S[i][j]-'a']=1;
		}cnt[S[i][0]-'a'][Has[i][1]]++;
	}cout<<ans<<"\n";
	return 0;
}