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++ (Clang 10.0.0) (judge verdict: Accepted)

#define LOCAL
#define _USE_MATH_DEFINES
#include <array>
#include <cassert>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <iomanip>
#include <string>
#include <sstream>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
#include <complex>
#include <cmath>
#include <numeric>
#include <bitset>
#include <functional>
#include <random>
#include <ctime>

using namespace std;

template <typename A, typename B>
ostream& operator <<(ostream& out, const pair<A, B>& a) {
  out << "(" << a.first << "," << a.second << ")";
  return out;
}
template <typename T, size_t N>
ostream& operator <<(ostream& out, const array<T, N>& a) {
  out << "["; bool first = true;
  for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
  return out;
}
template <typename T>
ostream& operator <<(ostream& out, const vector<T>& a) {
  out << "["; bool first = true;
  for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
  return out;
}
template <typename T, class Cmp>
ostream& operator <<(ostream& out, const set<T, Cmp>& a) {
  out << "{"; bool first = true;
  for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "}";
  return out;
}
template <typename U, typename T, class Cmp>
ostream& operator <<(ostream& out, const map<U, T, Cmp>& a) {
  out << "{"; bool first = true;
  for (auto& p : a) { out << (first ? "" : ", "); out << p.first << ":" << p.second; first = 0;} out << "}";
  return out;
}
#ifdef LOCAL
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...) 42
#endif
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}

typedef long long int64;
typedef pair<int, int> ii;
#define SZ(x) (int)((x).size())
const int INF = 1 << 29;
const int MOD = 1e9 + 7;
mt19937 mrand(random_device{}());
int rnd(int x) { return mrand() % x; }

struct fast_ios {
  fast_ios() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
  };
} fast_ios_;

const int M = 1e6 + 10;
char s[M];

struct Node {
  Node* ch[26];
  int cnt;
};
Node pool[M], *last;
Node* rt;

Node* new_node() {
  Node* cur = last++;
  fill(cur->ch, cur->ch + 26, nullptr);
  cur->cnt = 0;
  return cur;
}

void insert(char s[], int len, int target) {
  Node* p = rt;
  int pos;
  for (pos = len - 1; pos >= 0 && s[pos] - 'a' != target; --pos);
  for (int i = 0; i < len; ++i) {
    int k = s[i] - 'a';
    if (!p->ch[k]) p->ch[k] = new_node();
    p = p->ch[k];
    if (i <= pos) p->cnt += 1;
  }
  // if (pos >= 0) rt->cnt++;
}

int main() {
  int n;
  scanf("%d", &n);
  vector<int> offset(n + 1);
  int sum = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%s", s + sum);
    int len = strlen(s + sum);
    reverse(s + sum, s + sum + len);
    offset[i] = sum;
    sum += len;
  }
  offset[n] = sum;
  int64 ret = 0;
  for (int k = 0; k < 26; ++k) {
    last = pool;
    rt = new_node();
    for (int i = 0; i < n; ++i) {
      int len = offset[i + 1] - offset[i];
      insert(s + offset[i], len, k);
    }
    for (int i = 0; i < n; ++i) {
      int len = offset[i + 1] - offset[i];
      if (s[offset[i] + len - 1] - 'a' == k) {
        Node *p = rt;
        for (int j = 0; j < len - 1; ++j) {
          int x = s[offset[i] + j] - 'a';
          p = p->ch[x];
        }
        for (int k = 0; k < 26; ++k) {
          if (p->ch[k]) ret += p->ch[k]->cnt;
        }
        ret -= 1;
        // ret += p->cnt - 1;
      }
    }
    // trace(k, ret);
  }
  printf("%lld\n", ret);
  return 0;
}