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 PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

#b
import sys
def input(): return sys.stdin.readline().strip()
n = int(input())
s = [input() for _ in range(n)]
s.sort(key=len)
ans = 0
trie = {'!': True}
for si in s:
    cursor = trie
    si = list(si)
    while True:
        for nextchar in cursor:
            if nextchar == '!':
                continue
            if '!' in cursor[nextchar] and nextchar in si:
                ans += 1
        popchar = si.pop(-1)
        if popchar in cursor:
            cursor = cursor[popchar]
        else:
            cursor[popchar] = {}
            cursor = cursor[popchar]
            for popchar in reversed(si):
                cursor[popchar] = {}
                cursor = cursor[popchar]
            cursor['!'] = True
            break
print(ans)