Problem:
Score : 300 points

Problem Statement

There are N people. The name of the i-th person is S_i.

We would like to choose three people so that the following conditions are met:

The name of every chosen person begins with M, A, R, C or H.

There are no multiple people whose names begin with the same letter.

How many such ways are there to choose three people, disregarding order?

Note that the answer may not fit into a 32-bit integer type.

Constraints

1 \leq N \leq 10^5

S_i consists of uppercase English letters.

1 \leq |S_i| \leq 10

S_i \neq S_j (i \neq j)

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

If there are x ways to choose three people so that the given conditions are met, print x.

Sample Input 1

5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI

Sample Output 1

2

We can choose three people with the following names:

MASHIKE, RUMOI, HABORO

MASHIKE, RUMOI, HOROKANAI

Thus, we have two ways.

Sample Input 2

4
ZZ
ZZZ
Z
ZZZZZZZZZZ

Sample Output 2

0

Note that there may be no ways to choose three people so that the given conditions are met.

Sample Input 3

5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO

Sample Output 3

7

Solution in Kotlin (1.0.0) (judge verdict: Wrong Answer)

private val ACCEPTABLE_HEAD = setOf('M', 'A', 'R', 'C', 'H')

fun main(args: Array<String>) {
    val n = readLine()?.toInt() ?: return

    // 名前の頭文字とその数のmapを作ってしまう
    // MARCH以外の頭文字をもつ名前は除外する
    val sMap: Map<Char, Int> = (0 until n)
        .map { readLine()!![0] }
        .filter { it in ACCEPTABLE_HEAD }
        .groupBy { it }
        .mapValues { it.value.size }

    // sMapのsizeは今確実に5未満となっている
    // 5個から3つ選ぶ組み合わせを計算するために3重ループするが、
    // これは十分間に合う速度

    var count = 0L

    for (entry1 in sMap.entries) {
        for (entry2 in sMap.entries) {
            for (entry3 in sMap.entries) {
                // keyの順序を固定することで重複を排除する
                // Kotlin(1.0.0)だとcompareでChar to Numberの変換が行われてClassCastException
                // となるので、それを回避
                if (entry1.key.toInt() < entry2.key.toInt() &&
                    entry1.key.toInt() < entry3.key.toInt() &&
                    entry2.key.toInt() < entry3.key.toInt())
                {
                    count += (entry1.value * entry2.value * entry3.value)
                }
            }
        }
    }

    println(count)
}