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>
#include <unistd.h>
#include <sys/time.h>
#include <stdlib.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<P, ll> T;
typedef pair<long double , ll> Ps;
typedef pair<ll, bool> Pb;

const ll INF = 1e18;
const ll fact_table = 3200008;
long double Pi = 3.1415926535897932384626;

priority_queue <ll> pql;
priority_queue <P> pqp;
priority_queue <P> bag;
//big priority queue
priority_queue <ll, vector<ll>, greater<ll> > pqls;
priority_queue <P, vector<P>, greater<P> > pqps;
//small priority queue
//top pop

ll dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
ll dy[8] = {0, 1, 0, -1, 1, -1, -1, 1};
//↓,→,↑,←

/*
#define endl "\n"
#ifdef ENJAPMA
#undef endl
#endif
*/
#define p(x) cout<<x<<endl;
#define el cout<<endl;
#define pe(x) cout<<x<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<(x);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define CLEAR(a) a = decltype(a)();
#define pc(x) cout << x << ",";
#define rep(i, n) for (ll i = 0; i < (n); i ++)

// const ll mod = 998244353ll;
const ll mod = 1000000007ll;

ll mypow(ll a, ll b, ll mod) {ll x = 1; while (b) {while (!(b & 1)) {(a *= a) %= mod; b >>= 1;}(x *= a) %= mod; b--;} return x;}
void YES(bool cond) { if (cond) { p("YES");} else { p("NO");} return;}
void Yes(bool cond) { if (cond) { p("Yes");} else { p("No");} return;}
void line() { p("--------------------"); return;}

/*
ll fact[fact_table + 5], rfact[fact_table + 5];

void c3_init() {
    fact[0] = rfact[0] = 1;
    for (ll i = 1; i <= fact_table; i++) {
        fact[i] = (fact[i - 1] * i) % mod;
    }
    rfact[fact_table] = mypow(fact[fact_table], mod - 2, mod);
    for (ll i = fact_table; i >= 1; i--) {
        rfact[i - 1] = rfact[i] * i;
        rfact[i - 1] %= mod;
    }
    return;
}
ll c3(ll n, ll r) {
    return (((fact[n] * rfact[r]) % mod ) * rfact[n - r]) % mod;
}
*/
struct Timer {
    int64_t start;
    const int64_t CYCLES_PER_SEC = 2800000000;
    Timer() { reset(); }
    void reset() { start = getCycle(); }
    inline double get() { return (double)(getCycle() - start) / CYCLES_PER_SEC; }
    inline int64_t getCycle() {
        uint32_t low, high;
        __asm__ volatile ("rdtsc" : "=a" (low), "=d" (high));
        return ((int64_t)low) | ((int64_t)high << 32);
    }
};

Timer timer;

bool multicase = false;

ll n, m, k, w, num, a, b, c, d, e, h, q, ans, sum, l;

typedef vector<ll> vec;
typedef vector<vector<ll>> mat;

ll table[105][105][105];

struct LLRollingHash {
// long long に対するRollingHash
public:
    ll n;
    ll base;
    ll mod;
    vector<ll> hash_table;
    vector<ll> base_pow;

    LLRollingHash(vector<ll> v, ll my_base, ll my_mod) {
        n = v.size();
        base = my_base, mod = my_mod;
        hash_table.resize(n + 1, 0ll);
        base_pow.resize(n + 1, 1ll);
        ll tmp = 0;
        for(ll i=0;i<n;i++){
            assert(v[i] >= 0);
            tmp = tmp * base + v[i];
            tmp %= mod;
            hash_table[i+1] = tmp;
        }
        for(ll i=0;i<n;i++){
            base_pow[i+1] = (base_pow[i] * base) % mod;
        }

    }

    ll hash(ll len, ll idx){
        // idx番目の文字からn文字分のHash値を返す
        if(len == 0) return 0ll;
        assert(idx + len <= n);
        ll modpow = mod - base_pow[len];
        ll hash = (hash_table[len + idx] + (hash_table[idx] * modpow) % mod) % mod;
        return hash;
    }

    void add(ll val){
        n++;
        ll tmp = hash_table.back() * base + val;
        tmp %= mod;
        hash_table.pb(tmp);
        base_pow.pb((base_pow.back() * base) % mod);
        return;
    }
};

struct fin{
    ll len;
    ll hash;
    ll mozi;
    ll idx;
    
};

bool operator<(const fin& e1, const fin& e2) {
    if (e1.len != e2.len) {
        return e1.len < e2.len;
    }
    if (e1.hash != e2.hash) {
        return e1.hash < e2.hash;
    }
    if (e1.mozi != e2.mozi) {
        return e1.mozi < e2.mozi;
    }
    if (e1.idx != e2.idx) {
        return e1.idx < e2.idx;
    }
    return false;
}   

vector<LLRollingHash> x;
multiset<fin> st;
vector<ll> G[500005];
ll leng[200005];

bool exist[1000005];

string ss[200005];
typedef pair<ll, string> ls;
ls tableee[200005];

const ll mymod = 2000000011ll;



void solve(){
    cin >> n;

    for(int i=0;i<n;i++){
        cin >> tableee[i].second;
        tableee[i].first = tableee[i].second.size();
        exist[tableee[i].first] = true;
    }
    sort(tableee, tableee + n);
    reverse(tableee, tableee + n);
    for(int i=0;i<n;i++){
        string s;

        s = tableee[i].second;
        
        reverse(s.begin(), s.end());
        ll mylen = s.size() - 1;
        ll ret = 0;
        for(int j=0;j<s.size()-1;j++){
            ret *= 128ll;
            ret += s[j] - 'a' + 25;
            ret %= mymod;
        }

        ll myhash = ret;
        ret = 0;

        ll c = s.back() - 'a';

        auto ite2 = st.lower_bound(fin{mylen, myhash, c + 1, 0});
        auto ite1 = st.lower_bound(fin{mylen, myhash, c, (ll)s.size() - 1});
        /*
        p(s);
        line();
        p("search from");
        pe(mylen);pe(myhash);pe(c + 1);p(0);
        pe(mylen);pe(myhash);pe(c);p(s.size() - 1);
        line();

        for(auto e : st) {
            pe(e.len);pe(e.hash);pe(e.mozi);p(e.idx);
        }
        */
        ll idx1 = distance(st.begin(), ite1);
        ll idx2 = distance(st.begin(), ite2);


        ans += idx2 - idx1;

        vec table(26, -INF);
        for(int j=0;j<s.size();j++){
            table[s[j] - 'a'] = max(table[s[j] - 'a'], (ll)j);
        }
        for(int k=0;k<26;k++){
            if (table[k] != -INF) st.insert(fin{0, 0, k, table[k]});
        }
        for(int j=0;j<s.size();j++){
            ret *= 128ll;
            ret += s[j] - 'a' + 25;
            ret %= mymod;
            if (!exist[j + 2]) continue;
            for(int k=0;k<26;k++){
                if (table[k] != -INF) st.insert(fin{j + 1, ret, k, table[k]});
            }
        }


    }
    p(ans);
    return;
}

int main() {
    // init();
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll q, testcase = 1;
    if (multicase) {
        cin >> q;
    } else {
        q = 1;
    }
    while (q--) {
        // pu("Case ");pu("#");pu(testcase);pu(": ");
        solve();
        testcase++;
    }
    // solve();
    return 0;
}