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)

import sys
sys.setrecursionlimit(10**7) #再帰関数の上限,10**5以上の場合python
import math
from copy import copy, deepcopy
from copy import deepcopy as dcp
from operator import itemgetter
from bisect import bisect_left, bisect, bisect_right#2分探索
#bisect_left(l,x), bisect(l,x)#aはソート済みである必要あり。aの中からx未満の要素数を返す。rightだと以下
from collections import deque, defaultdict
#deque(l), pop(), append(x), popleft(), appendleft(x)
#q.rotate(n)で → にn回ローテート
from collections import Counter#文字列を個数カウント辞書に、
#S=Counter(l),S.most_common(x),S.keys(),S.values(),S.items()
from itertools import accumulate,combinations,permutations,product#累積和
#list(accumulate(l))
from heapq import heapify,heappop,heappush
#heapify(q),heappush(q,a),heappop(q) #q=heapify(q)としないこと、返り値はNone
from functools import reduce,lru_cache#pypyでもうごく
#@lru_cache(maxsize = None)#maxsizeは保存するデータ数の最大値、2**nが最も高効率
from decimal import Decimal

def input(): 
    x=sys.stdin.readline()
    return x[:-1] if x[-1]=="\n" else x
def printe(*x):print("## ",*x,file=sys.stderr)
def printl(li): _=print(*li, sep="\n") if li else None
def argsort(s, return_sorted=False): 
    inds=sorted(range(len(s)), key=lambda k: s[k])
    if return_sorted: return inds, [s[i] for i in inds]
    return inds
def alp2num(c,cap=False): return ord(c)-97 if not cap else ord(c)-65
def num2alp(i,cap=False): return chr(i+97) if not cap else chr(i+65)
def matmat(A,B):
    K,N,M=len(B),len(A),len(B[0])
    return [[sum([(A[i][k]*B[k][j]) for k in range(K)]) for j in range(M)] for i in range(N)]
def matvec(M,v):
    N,size=len(v),len(M)
    return [sum([M[i][j]*v[j] for j in range(N)]) for i in range(size)]
def T(M):
    n,m=len(M),len(M[0])
    return [[M[j][i] for j in range(n)] for i in range(m)]
def binr(x): return bin(x)[2:]
def bitcount(x): #xは64bit整数
    x= x - ((x >> 1) & 0x5555555555555555)
    x= (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
    x= (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f 
    x+= (x >> 8); x+= (x >> 16); x+= (x >> 32) 
    return x & 0x7f

def main():
    mod = 1000000007
    #w.sort(key=itemgetter(1),reverse=True)  #二個目の要素で降順並び替え

    N = int(input())
    #N, K = map(int, input().split())
    #A = tuple(map(int, input().split())) #1行ベクトル
    L=["" for _ in range(N)]
    ll=[0]*N
    for i in range(N):
        s=input()
        ls=len(s)
        ll[i]=ls
        L[i]=s

    #S = tuple(tuple(map(int, input().split())) for i in range(N)) #改行行列

    inds=argsort(ll)
    d=defaultdict(Counter)
    ans=0
    for im in inds:
        s=L[im]

        ls=len(s)
        cur=""
        se=Counter(s)
        #printe(se)
        for i in range(ls):
            for a,val in se.items():
                if val:
                    ans+=d[a][cur]

            si=s[-1-i]
            cur+=si
            

            se[si]-=1
        a=cur[-1]
        d[a][cur[0:-1]]+=1

    print(ans)
            





if __name__ == "__main__":
    main()