shil-and-lcp-pairsmonk
Source: hackerearth
Difficulty: MEDIUM_HARD

Shil came across N strings - s1 , s2 , s3 .. sN . He wants to find out total number of unordered pairs (i, j) such that LCP(si , sj) = k. You have to do thi for every k in [0, L].  Here L = max ( len(si) ) and  LCP denotes longest common prefix.

INPUT:
First line consists of integer N denoting total number of strings.Next N lines consists of strings  s1 , s2 , s3 .. sN.  

OUTPUT:
Print L+1 integers  - h0 , h1 , h2, ..., hL where hi denotes total number of pairs having LCP value i and L = max ( len(si) ).

CONTRAINTS:
1 ≤ N ≤ 10^5
1≤ Σ len(si) ≤ 10^6
Strings consists of lower case english alphabets ( 'a' - 'z' ) only.

SAMPLE INPUT
5
aaaa
ab
abcd
aaae
g

SAMPLE OUTPUT
4 4 1 1 0 

Explanation

All the pairs having LCP 0 are (1,5) , (2,5) , (3,5) , (4,5)
All the pairs having LCP 1 are (1,2) , (1,3) , (2,4) , (3,4)
All the pairs having LCP 3 are (1,4)
All the pairs having LCP 2 are (2, 3)