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)

#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
#define ll long long
#define ld long double
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repo(i,n) for(int i = 1; i < (int)(n); i++)
#define pb push_back
#define mp make_pair
#define np next_permutation
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define uniq(v) v.erase(unique(v.begin(),v.end()),v.end())
#define lb(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define ub(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
using Pair = pair<ll,pair<int,int>>;
#define pq priority_queue<Pair, vector<Pair>, greater<Pair>> 
const ll mod=1000000007;
//const ll mod=998244353;
const ld pi=acos(-1.0);
const ll INF = 1LL<<61;
template<class T>bool chmax(T &a, const T &b) { 
  if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) {
  if (b<a) { a=b; return 1; } return 0; }
ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }
//intの最大値2147483647 ≒ 2×10^9
//long longの最大値9223372036854775807 ≒ 9×10^18
//'大文字'+=32;    で小文字に
//  cout << fixed << setprecision (20);   小数点以下2０桁まで
//実行時間制約2秒では２×10^8回くらいまで計算できる



int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin>>n;

  int a=0;
  int b=0;
  int c=0;
  int d=0;
  rep(i,n){
    string s;
    cin>>s;
    if(s[0]=='A') a++;
    if(s[0]=='W') b++;
    if(s[0]=='T') c++;
    if(s[0]=='R') d++;
  }

  cout << "AC x " << a << endl;
  cout << "WA x " << b << endl;
  cout << "TLE x " << c << endl;
  cout << "RE x " << d << endl;

}