Problem:
Score : 200 points

Problem Statement

Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A.

The problem has N test cases.

For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is AC, WA, TLE, and RE, respectively.

See the Output section for the output format.

Constraints

1 \leq N \leq 10^5

S_i is AC, WA, TLE, or RE.

Input

Input is given from Standard Input in the following format:

N
S_1
\vdots
S_N

Output

Let C_0, C_1, C_2, and C_3 be the numbers of test cases for which the verdict is AC, WA, TLE, and RE, respectively. Print the following:

AC x C_0
WA x C_1
TLE x C_2
RE x C_3

Sample Input 1

6
AC
TLE
AC
AC
WA
TLE

Sample Output 1

AC x 3
WA x 1
TLE x 2
RE x 0

We have 3, 1, 2, and 0 test case(s) for which the verdict is AC, WA, TLE, and RE, respectively.

Sample Input 2

10
AC
AC
AC
AC
AC
AC
AC
AC
AC
AC

Sample Output 2

AC x 10
WA x 0
TLE x 0
RE x 0

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

//Let's join Kaede Takagaki Fan Club !!
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
#define all(x) x.begin(),x.end()
template<class T>
void dmp(T a){
	rep(i,a.size()) cout << a[i] << " ";
	cout << endl;
}
template<class T>
bool chmax(T&a, T b){
	if(a < b){
		a = b;
		return 1;
	}
	return 0;
}
template<class T>
bool chmin(T&a, T b){
	if(a > b){
		a = b;
		return 1;
	}
	return 0;
}
template<class T>
void g(T &a){
	cin >> a;
}
template<class T>
void o(const T &a,bool space=false){
	cout << a << (space?' ':'\n');
}
//ios::sync_with_stdio(false);
const ll mod = 1000000007;//998244353
template<class T>
void add(T&a,T b){
	a+=b;
	if(a >= mod) a-=mod;
}
struct UF{
	int par[2000005],ran[2000005];
	void init(){ for(int i=0;i<2000005;i++) par[i] = i, ran[i] = 0; }
	int find(int x){ if(x == par[x]) return x; else return par[x] = find(par[x]); }
	void unite(int x,int y){
		x = find(x); y = find(y); if(x==y) return;
		if(ran[x] < ran[y]) par[x] = y;
		else{
			par[y] = x;
			if(ran[x] == ran[y]) ran[x]++;
		}
	}
	bool same(int x,int y){ return find(x)==find(y); }
}uf;


ll modpow(ll x,ll n){
	ll res=1;
	while(n>0){
		if(n&1) res=res*x%mod;
		x=x*x%mod;
		n>>=1;
	}
	return res;
}
ll F[2000005],R[2000005];
void make(){
	F[0] = 1;
	for(int i=1;i<2000005;i++) F[i] = F[i-1]*i%mod;
	R[2000004] = modpow(F[2000004], mod-2);
	for(int i=2000003;i>=0;i--) R[i] = R[i+1] * (i+1) % mod;
}
ll C(int a,int b){
	return F[a]*R[b]%mod*R[a-b]%mod;
}
void solve(){
	int n; string s;
	map<char, int>M;
	cin >> n;
	rep(i, n){
		cin >> s;
		M[s[0]] ++;
	}
	cout << "AC x " << M['A'] << endl;
	cout << "WA x " << M['W'] << endl;
	cout << "TLE x " << M['T'] << endl;
	cout << "RE x " << M['R'] << endl;
}




int main(){
	uf.init();
	make();
	solve();
}