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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define rep(i, n) for (ll i = 0; i < n; i++)
#define FOR(i, a, b) for (ll i = a; i < b; i++)
#define is(a, b) a == b
#define len(v) ll(v.size())

class UnionFind {
public:
	vector<int> par;//親の番号
    vector<int> rank;//木の深さ
 
	UnionFind(int n){//全ての要素に対して自身が根となるように初期化, 木の深さは0になる
		rep(i,n){
            par.push_back(i);
            rank.push_back(0);
        }
	}
 
	int find(int x) {//木の根を求める
		if(par[x]==x){
            return x;
        }else{
            return par[x]=find(par[x]);
        }
	}
 
	void unite(int x,int y){//xとyの属する集合を併合
        x=find(x);
        y=find(y);
        if(x==y) return;
        if(rank[x]<rank[y]){//木の深さが小さい方を大きい方へ繋げる
            par[x]=y;
        }else{
            par[y]=x;
            if(rank[x]==rank[y]) rank[x]++;
        }
    }

    bool same(int x,int y){//xとyが同じ集合に属するか否か
        return find(x)==find(y);
    }
};
//x以下の要素への最小index, ソートしてから使ってね
template <class T> 
int former(const vector<T> &v, T x){
  return upper_bound(v.begin(), v.end(), x) - v.begin() - 1;
}
//x以上の要素への最小index
template <class T> 
int latter(const vector<T> &v, T x){
  return lower_bound(v.begin(), v.end(), x) - v.begin();
}
//vector書き出し
template <class T>
void cout_vec(const vector<T> &vec1){
  rep(i,len(vec1)){
    cout<<vec1[i]<<' ';
  }
  cout<<'\n';
}
//nCk
template <class T>
T comb(T n,T k){
    T ans=1;
    FOR(i,n-k+1,n+1){
        ans*=i;
    }
    FOR(i,1,k+1){
        ans/=i;
    }
    return ans;
}
//a,bの最大公約数
template <class T>
T gcd(T a,T b) {
	if (b == 0) return a;
	else return gcd(b, a % b);
}
//素因数分解
template <class T>
map<T,T> prime_factor(T x){
    map<T,T> ans;
    for(T i=2;i*i<=x;i++){
        while(x%i==0){
            x/=i;
            ans[i]++;
        }
    }
    if (x!=1){
        ans[x]=1;
    }
    return ans;
}
//約数列挙
template <class T>
vector<T> divisor(T x){
    vector<T> res;
    for(T i=1;i*i<=x;i++){
        if(x%i==0){
            res.push_back(i);
            if(i!=x/i){
                res.push_back(x/i);
            }
        }
    }
    return res;
}
//素数判定
template <class T>
bool is_prime(T x){
    for(T i=2;i*i<=x;i++){
        if(x%i==0){
            return false;
        }
    }
    return x!=1;
}

const ll inf=1e9+7;

ll mult(ll n,ll m){//n^m%infを求める
    if(m==1){
        return n%inf;
    }else if(m%2==0){//オーバーフロー防止のためにn^(m/2)*n^(m/2)を考える
        ll t=mult(n,m/2);
        return (t*t)%inf;
    }else{//奇数の時はn^(m-1)*nを考える
        ll t=mult(n,m-1);
        return (t*n)%inf;
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    map<string,ll> s;
    string x;
    rep(i,n){
        cin>>x;
        if(x[0]=='M'){
            s["M"]++;
        }else if(x[0]=='A'){
            s["A"]++;
        }else if(x[0]=='R'){
            s["R"]++;
        }else if(x[0]=='C'){
            s["C"]++;
        }else if(x[0]=='H'){
            s["H"]++;
        }
    }
    ll ans=s["M"]*s["A"]*s["R"]+s["M"]*s["A"]*s["C"]+s["M"]*s["A"]*s["H"]+s["M"]*s["R"]*s["C"]+s["M"]*s["R"]*s["H"]+s["M"]*s["C"]*s["H"]+s["A"]*s["R"]*s["C"]+s["A"]*s["R"]*s["H"]+s["A"]*s["C"]*s["H"]+s["R"]*s["C"]*s["H"];
    cout<<ans<<endl;
}