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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

fun main() {
  val n = readLine()!!.toInt()
  val s = Array(n) { readLine()!!.reversed() }
  s.sort()
  val trie = Trie()
  var x = 0L
  for (si in s) {
    x += trie.count(si)
    trie.insert(si)
  }
  println(x)
}
class Trie {
  private var size = 0
  private val childs = Array<Trie?>(26) { null }

  fun insert(s: String) {
    var node = this
    for (c in s) {
      val n = c - 'a'
      if (node.childs[n] == null) node.childs[n] = Trie()
      node = node.childs[n]!!
    }
    node.size++
  }

  fun count(s: String): Long {
    val cnt = IntArray(26)
    for (c in s) {
      val n = c - 'a'
      cnt[n]++
    }
    var res = 0L
    var node = this
    for (c in s) {
      for (i in cnt.indices) {
        if (cnt[i] > 0 && node.childs[i] != null) {
          res += node.childs[i]!!.size
        }
      }
      val n = c - 'a'
      if (node.childs[n] == null) break
      node = node.childs[n]!!
    }
    return res
  }
}