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;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ll long long 
#define REP(i, a, b) for (ll i = a; i < b; i++)
#define REPI(i, a, b) for (ll i = b - 1; i >= a; i--)
#define i_os ios::sync_with_stdio(0);  cin.tie(0);  cout.tie(0);
#define INF (ll)1e18 + 100
#define endl "\n"
#define p0(a) cout << a << " "
#define p1(a) cout << a << endl
#define p2(a, b) cout << a << " " << b << endl
#define p3(a, b, c) cout << a << " " << b << " " << c << endl
#define p4(a, b, c, d) cout << a << " " << b << " " << c << " " << d << endl
/*SOME BITMASK KNOWLEDGE
1)x & (x - 1):sets the last one bit of x to zero
power of two exactly when x & (x − 1) = 0.
2)x & -x:sets all the one bits to zero, except last one bit
3)x | (x - 1):inverts all the bits after the last one bit*/
#define o_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define o_setll tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>
typedef tree<pair<ll, ll>,null_type,less<pair<ll, ll>>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> indexed_set1;
typedef tree<string,null_type,less<string>,rb_tree_tag,tree_order_statistics_node_update> indexed_set2;
//1. order_of_key(k) : number of elements strictly lesser than k
//2. find_by_order(k) : k-th element in the set
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);

bool compare(string s1,string s2){
    return s1.size() > s2.size();
}
int main()
{
    i_os;
    ll n;
    cin>>n;
    string arr[n];
    REP(i,0,n){
        cin>>arr[i];
    }
    sort(arr,arr+n,compare);

    ll ct = 0;
    map<pair<string,char>,ll> mp;
    REP(i,0,n){
        string p1 = arr[i].substr(1);
        if(mp.count({p1,arr[i][0]})){
            ct += mp[{p1,arr[i][0]}];
        }

        ll temp[26] = {0};
        REP(j,0,arr[i].size()){
            temp[arr[i][j]-'a']++;
        }
        REP(k,0,26){
            if(temp[k] > 0){
                mp[{"",k+'a'}]++;
            }
        }
        string s = "";
        REPI(j,0,arr[i].size()){
            temp[arr[i][j]-'a']--;
            s = arr[i][j] + s;
            REP(k,0,26){
                if(temp[k] > 0){
                    mp[{s,k+'a'}]++;
                }
            }
        }
    }
    cout<<ct<<endl;
    return 0;
}