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)

from collections import Counter, defaultdict as dd
N = int(input())

ans = 0
Ss = []
orda = ord('a')
for _ in range(N):
    S = input()
    Ss.append((len(S), list(map(lambda c:ord(c)-orda, S))))
Ss.sort()

ctkey = 26
RTrie = [0]*27
for i in range(N):
    _, S = Ss.pop()
    charct = Counter(S)
    node = RTrie
    L = len(S)
    for i in range(L-1, -1, -1):
        c = S[i]
        for a in charct.keys():
            if not node[a]:
                node[a] = [0]*27
            node[a][ctkey] += 1
        charct[c] -= 1
        if charct[c] == 0:
            charct.pop(c)
        node = node[c]
    ans += node[ctkey]-1

print(ans)