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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define all(a) a.begin(), a.end()
#define pb push_back
#define mp make_pair
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<ll,int> pli;
typedef pair<int,int> pii;
#define rep(i,a,b) for(ll i=a ; i<b ; i++)
#define qrep(que, ite) for(auto ite=begin(que) ; ite!=end(que) ; ite++)
const int max_n = 1e5;
const ll mod = 1e9+7;
const ll INF = 1e17;
const int inf = 1e5;
typedef long double ld;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
ll gcd(ll a, ll b) { return a ? gcd(b%a, a) : b; }
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
ll mo = 998244353;
struct UnionFind {
    vector<int> par;

    UnionFind(int n) : par(n, -1) { }
    void init(int n) { par.assign(n, -1); }

    int root(int x) {
        if (par[x] < 0) return x;
        else return par[x] = root(par[x]);
    }

    bool issame(int x, int y) {
        return root(x) == root(y);
    }

    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (par[x] > par[y]) swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }

    int size(int x) {
        return -par[root(x)];
    }
};
struct SegmentTree{
  int N;
  vector<int> node;
  public :
    void intit(vector<int>v){
      int sz = v.size();
      N=1;
      while(N<sz) N*=2;
      node.resize(N);
      for(int i=0 ; i<sz ; i++) node[i+N-1] = v[i];
      for(int i=N-2 ; i>=0 ; i--) node[i] = min(node[i*2+1], node[i*2+2]);
    }
    void update(int x, int val){
      x += N-1;
      node[x+N-1] = val;
      while(x>0){
        x = (x-1)/2;
        node[x] = min(node[x*2+1], node[x*2+2]);
      }
    }
    int getmin(int a, int b, int k, int l, int r){
      if(b<=l || r<=a) return inf;
      else if(a<=l && r<=b) return node[k];
      else{
        int vl = getmin(a, b, 2*k+1, l, (l+r)/2);
        int vr = getmin(a, b, 2*k+2, (l+r)/2, r);
        return min(vl, vr);
      }
    }
};
vector<ll> divisor(ll n){
  vector<ll> res;
  for(ll i=1; i*i<=n ; i++){
    while(n%i==0){
      ++res[i];
      n /= i;
    }
  }
  if(n!=1) res[n]=1;
  return res;
}
int main(){
  int N;  cin>>N;
  vector<string> v(N);
  rep(i,0,N) cin>>v[i];
  int a=0,b=0,c=0,d=0;
  rep(i,0,N){
    if(v[i]=="AC") a++;
    else if(v[i]=="WA") b++;
    else if(v[i]=="TLE") c++;
    else c++;
  }
  cout<<"AC x "<<a<<endl;;
  cout<<"WA x "<<b<<endl;;
  cout<<"TLE x "<<c<<endl;;
  cout<<"RE x "<<d<<endl;;
  return 0;
}