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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define REP(i, start, end) for (ll i = start, i##Len = (end); i < i##Len; ++i)
#define REPR(i, start, end) for (ll i = start, i##Len = (end); i > i##Len; --i)
using ll = long long;
using namespace std;

struct Node
{
    int end_count = 0;
    bool init = 0;
    vector<Node> nodes;

    void add(string &s)
    {
        if (s.empty())
            end_count++;
        else
        {
            if (!init)
            {
                nodes.assign(26, Node{});
                init = 1;
            }
            int idx = s.back() - 'a';
            s.pop_back();
            nodes[idx].add(s);
        }
    }
};

struct Trie
{
    Node node = Node{};

    void add(string &s)
    {
        node.add(s);
    }
};

int main()
{
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);

    int N;
    cin >> N;
    vector<string> a(N);
    REP(i, 0, N)
    cin >> a[i];
    sort(a.begin(), a.end(), [](string a, string b) { return a.size() < b.size(); });
    auto trie = Trie{};
    ll ans = 0;

    for (auto &w : a)
    {
        vector<int> cnt(26);
        for (auto &c : w)
            cnt[c - 'a']++;

        Node *node = &trie.node;
        REPR(i, w.size() - 1, -1)
        {
            if (!node->init)
            {
                break;
            }
            REP(j, 0, 26)
            {
                if (cnt[j] > 0)
                    ans += node->nodes[j].end_count;
            }
            cnt[w[i] - 'a']--;
            node = &node->nodes[w[i] - 'a'];
        }

        trie.add(w);
    }

    cout << ans << endl;
}