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 Python (3.4.3) (judge verdict: Accepted)

###############################################################################

from sys import stdout
from bisect import bisect_left as binl
from copy import copy, deepcopy
from collections import defaultdict


mod = 1


def intin():
    input_tuple = input().split()
    if len(input_tuple) <= 1:
        return int(input_tuple[0])
    return tuple(map(int, input_tuple))


def intina():
    return [int(i) for i in input().split()]


def intinl(count):
    return [intin() for _ in range(count)]


def modadd(x, y):
    global mod
    return (x + y) % mod


def modmlt(x, y):
    global mod
    return (x * y) % mod


def lcm(x, y):
    while y != 0:
        z = x % y
        x = y
        y = z
    return x


def combination(x, y):
    assert(x >= y)

    if y > x // 2:
        y = x - y

    ret = 1

    for i in range(0, y):
        j = x - i
        i = i + 1

        ret = ret * j
        ret = ret // i

    return ret


def get_divisors(x):
    retlist = []
    for i in range(1, int(x**0.5) + 3):
        if x % i == 0:
            retlist.append(i)
            retlist.append(x // i)
    return retlist


def get_factors(x):
    retlist = []
    for i in range(2, int(x**0.5) + 3):
        while x % i == 0:
            retlist.append(i)
            x = x // i
    retlist.append(x)
    return retlist


def make_linklist(xylist):
    linklist = {}
    for a, b in xylist:
        linklist.setdefault(a, [])
        linklist.setdefault(b, [])
        linklist[a].append(b)
        linklist[b].append(a)
    return linklist


def calc_longest_distance(linklist, v=1):
    distance_list = {}
    distance_count = 0
    distance = 0
    vlist_previous = []
    vlist = [v]
    nodecount = len(linklist)

    while distance_count < nodecount:
        vlist_next = []
        for v in vlist:
            distance_list[v] = distance
            distance_count += 1
            vlist_next.extend(linklist[v])
        distance += 1
        vlist_to_del = vlist_previous
        vlist_previous = vlist
        vlist = list(set(vlist_next) - set(vlist_to_del))

    max_distance = -1
    max_v = None
    for v, distance in distance_list.items():
        if distance > max_distance:
            max_distance = distance
            max_v = v

    return (max_distance, max_v)


def calc_tree_diameter(linklist, v=1):
    _, u = calc_longest_distance(linklist, v)
    distance, _ = calc_longest_distance(linklist, u)
    return distance


class UnionFind:

    def __init__(self, n):
        self.parent = [i for i in range(n)]

    def root(self, i):
        if self.parent[i] == i:
            return i
        self.parent[i] = self.root(self.parent[i])
        return self.parent[i]

    def unite(self, i, j):
        rooti = self.root(i)
        rootj = self.root(j)
        if rooti == rootj:
            return
        if rooti < rootj:
            self.parent[rootj] = rooti
        else:
            self.parent[rooti] = rootj

    def same(self, i, j):
        return self.root(i) == self.root(j)


###############################################################################


def main():
    n = intin()
    slist = [input() for _ in range(n)]

    count = {'M': 0, 'A': 0, 'R': 0, 'C': 0, 'H': 0}

    for s in slist:
        if s[0] in count:
            count[s[0]] += 1

    patterns = [
        ['M', 'A', 'R'],
        ['M', 'A', 'C'],
        ['M', 'A', 'H'],
        ['M', 'R', 'C'],
        ['M', 'R', 'H'],
        ['M', 'C', 'H'],
        ['A', 'R', 'C'],
        ['A', 'R', 'H'],
        ['A', 'C', 'H'],
        ['R', 'C', 'H'],
    ]

    ans = 0

    for p in patterns:
        ans += count[p[0]] * count[p[1]] * count[p[2]]

    print(ans)


if __name__ == '__main__':
    main()