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>
using namespace std;


#define li          long long int
#define uli         unsigned long long int
#define rep(i,to)   for(li i=0;i<((li)(to));i++)
#define repp(i,start,to)    for(li i=(li)(start);i<((li)(to));i++)
#define pb          push_back
#define sz(v)       ((li)(v).size())
#define bgn(v)      ((v).begin())
#define eend(v)     ((v).end())
#define allof(v)    (v).begin(), (v).end()
#define dodp(v,n)       memset(v,(li)n,sizeof(v))
#define bit(n)      (1ll<<(li)(n))
#define mp(a,b)     make_pair(a,b)
#define rin rep(i,n)
#define EPS 1e-12
#define ETOL 1e-8
#define MOD 1000000007
typedef pair<li, li> PI;

#define INF bit(60)

#define DBGP 1


#define idp if(DBGP)
#define F first
#define S second
#define p2(a,b)     idp cout<<a<<"\t"<<b<<endl
#define p3(a,b,c)       idp cout<<a<<"\t"<<b<<"\t"<<c<<endl
#define p4(a,b,c,d)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<endl
#define p5(a,b,c,d,e)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<endl
#define p6(a,b,c,d,e,f)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<endl
#define p7(a,b,c,d,e,f,g)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<endl
#define p8(a,b,c,d,e,f,g,h)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<endl
#define p9(a,b,c,d,e,f,g,h,i)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<"\t"<<i<<endl
#define p10(a,b,c,d,e,f,g,h,i,j)        idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<"\t"<<i<<"\t"<<j<<endl
#define foreach(it,v)   for(__typeof((v).begin()) it=(v).begin(); it!=(v).end(); ++it)
#define p2p(x)      idp p2((x).F, (x).S)
#define dump(x,n)   idp{rep(i,n){cout<<x[i]<<" ";}puts("");}
#define dump2(x,n)  idp{rep(i,n){cout<<"["<<x[i].F<<" , "<<x[i].S<<"] ";}puts("");}
#define dumpi(x)    idp{foreach(it, x){cout<<(*it)<<" ";}puts("");}
#define dumpi2(x)   idp{foreach(it, x){cout<<"["<<(it)->F<<" , "<<(it)->S<<"] ";}puts("");}

#define read2d(a,w,h)   rep(i,h)rep(j,w)cin>>a[i][j]
#define dump2d(a,w,h)   rep(i,h){rep(j,w)cout<<a[i][j]<<" ";puts("");}

typedef pair<li, li> PI;

string s[200200];
li n;

namespace std {
template <>
class hash<std::pair<uli, uli>> {
public:
    size_t operator()(const std::pair<uli, uli>& x) const {
        return hash<uli>()(x.first) ^ hash<uli>()(x.second);
    }
};
}


unordered_map<uli, li> cnt[26];
unordered_map<pair<uli, uli>, li> cnt2[26];

uli hash_plus(uli cumul, uli add, uli base) {
    return cumul * base + add;
}

int main() {
    cin >> n;
    rin{
        cin >> s[i];
    }
    sort(s, s + n, [](const string & a, const string & b) {
        if (sz(a) != sz(b)) {
            return sz(a) < sz(b);
        }
        return a < b;
    });

    li res = 0;
    rin{
        set<li> st;
        vector<uli> hashes1(sz(s[i]) + 1);
        vector<uli> hashes2(sz(s[i]) + 1);
        hashes1[0] = 0;
        hashes2[0] = 0;
        rep(j, sz(s[i])) {
            hashes1[j + 1] = hash_plus(hashes1[j], s[i][sz(s[i]) - 1 - j] - 'a', 29);
            hashes2[j + 1] = hash_plus(hashes2[j], 26 - (s[i][sz(s[i]) - 1 - j] - 'a'), 31);
        }
        rep(j, sz(s[i])) {
            st.insert(s[i][j] - 'a');
            pair<uli, uli> h = {hashes1[sz(s[i]) - j - 1], hashes2[sz(s[i]) - j - 1]};
            for (auto v : st) {
                if (cnt2[v].count(h) > 0) {
                    res += cnt2[v][h];
                }
            }
        }
        cnt2[s[i][0] - 'a'][{hashes1[sz(s[i]) - 1], hashes2[sz(s[i]) - 1]}]++;
    }
    cout << res << endl;

    return 0;
}