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)

/*input
6
b
a
abc
c
d
ab
*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<pair<int,int>, null_type, less<pair<int,int>>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
typedef long long int ll;
#define ld double
#define pii pair<ll,ll>
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
const ll maxn=2e5+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=1000004099;
const ld PI=3.14159265358979323846;
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define MP make_pair
ll mult(ll a,ll b){
    ll res=0LL;
    while(b){
        if(b&1) res=(res+a)%MOD;
        a=(a+a)%MOD;
        b>>=1;
    }
    return res%MOD;
}
ll mypow(ll a,ll b){
    ll res=1LL;
    while(b){
        if(b&1) res=res*a%MOD;
        a=a*a%MOD;
        b>>=1;
    }
    return res;
}
string arr[maxn];
unordered_map<int,int> mp;
int cnt[26];
int32_t main(){
    ios::sync_with_stdio(false),cin.tie(0);
mp.reserve(32768);
mp.max_load_factor(0.25);
    int n;
    cin>>n;
    ll ans=0;
    ll inv=(mypow(26,MOD-2));
    REP(i,n){
        cin>>arr[i];
        string s=arr[i];
        reverse(ALL(s));
        ll hsh=0;
        REP(j,sz(s)){
            hsh=(hsh*26LL+(s[j]-'a'));
            if(hsh>=MOD) hsh%=MOD;
        }
        mp[hsh]++;
    }
    REP(i,n){
        string s=arr[i];
        string suff="";
        REP(j,sz(s)) cnt[s[j]-'a']++;
        ll hsh=0;
        for(int j=sz(s)-1;j>=0;j--){
            REP(k,26){
                if(cnt[k]){
                    hsh=(hsh*26LL+k);
                    if(hsh>=MOD) hsh%=MOD;
                    ans+=mp[hsh];
                    hsh-=k;
                    if(hsh<0){
                        hsh+=MOD;
                        hsh%=MOD;
                    }
                    hsh*=inv;
                    if(hsh>=MOD) hsh%=MOD;
                }
            }
            cnt[s[j]-'a']--;
            hsh=(hsh*26LL+(s[j]-'a'));
            if(hsh>=MOD) hsh%=MOD;
        }
        REP(j,26) cnt[j]=0;
    }   
    cout<<ans-n;
}