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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

// 内部定数
#define D_CHAR_KIND		5										// 文字種類

// 内部変数
static long long sl1Cnt[D_CHAR_KIND];							// 各文字毎の人数
static long long slSum;											// 合計

// 内部変数 - テスト用
#ifdef D_TEST
	static int siRes;
	static FILE *szpFpT, *szpFpA;
#endif

// 実行メイン
int
fMain(
	int piTNo					// <I> テスト番号 1～
)
{
	int i, j, k;
	char lc1Buf[1024], lc1Out[1024];

	// データ初期化
	memset(sl1Cnt, 0, sizeof(sl1Cnt));							// 各文字毎の人数
	slSum = 0;													// 合計

	// テストファイルオープン
#ifdef D_TEST
	sprintf(lc1Buf, ".\\Test\\T%d.txt", piTNo);
	szpFpT = fopen(lc1Buf, "r");
	sprintf(lc1Buf, ".\\Test\\A%d.txt", piTNo);
	szpFpA = fopen(lc1Buf, "r");
	siRes = 0;
#endif

	// 人数取得
	int liHCnt;
#ifdef D_TEST
	fgets(lc1Buf, sizeof(lc1Buf), szpFpT);
#else
	fgets(lc1Buf, sizeof(lc1Buf), stdin);
#endif
	sscanf(lc1Buf, "%d", &liHCnt);

	// 各文字毎の人数セット
	for (i = 0; i < liHCnt; i++) {
#ifdef D_TEST
		fgets(lc1Buf, sizeof(lc1Buf), szpFpT);
#else
		fgets(lc1Buf, sizeof(lc1Buf), stdin);
#endif

		switch (lc1Buf[0]) {
			case 'M':	sl1Cnt[0]++;	break;
			case 'A':	sl1Cnt[1]++;	break;
			case 'R':	sl1Cnt[2]++;	break;
			case 'C':	sl1Cnt[3]++;	break;
			case 'H':	sl1Cnt[4]++;	break;
		}
	}

	// 合計セット
	for (i = 0; i < D_CHAR_KIND; i++) {
		for (j = i + 1; j < D_CHAR_KIND; j++) {
			for (k = j + 1; k < D_CHAR_KIND; k++) {
				slSum += sl1Cnt[i] * sl1Cnt[j] * sl1Cnt[k];
			}
		}
	}

	// 結果セット
	sprintf(lc1Out, "%lld\n", slSum);

	// 結果表示
#ifdef D_TEST
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, lc1Out)) {
		siRes = -1;
	}
#else
	printf("%s", lc1Out);
#endif

	// テストファイルクローズ
#ifdef D_TEST
	fclose(szpFpT);
	fclose(szpFpA);
#endif

	// テスト結果
#ifdef D_TEST
	if (siRes == 0) {
		printf("OK %d\n", piTNo);
	}
	else {
		printf("NG %d\n", piTNo);
	}
#endif

	return 0;
}

int
main()
{

#ifdef D_TEST
	int i;
	for (i = D_TEST_SNO; i <= D_TEST_ENO; i++) {
		fMain(i);
	}
#else
	fMain(0);
#endif

	return 0;
}