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)

///Code by: Luis Alejandro
///Handle: Sekai
///Country: Cuba

#include <bits/stdc++.h>
#define endl '\n'
#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline")
#pragma GCC option("arch=native","tune=native","no-zero-upper")
#pragma GCC target("avx2")

using namespace std;

typedef unsigned long long ull;
typedef long long lld;
typedef long double ld;
typedef pair<int,int> pii;

const int MAXN=1e5+5;
const int MOD=1e9+7;
const int INF=INT_MAX;
const ld PI=acos(-1);
const ld EPS=1e-9;

#define lg2(x) 31-__builtin_clz(x)
#define logx(a,b) (log(a)/log(b))
#define read(x) cin>>x;

pii mkp(int a, int b){
    if(a>b)swap(a,b);
    return make_pair(a,b);
}

int mod_pow(int a, int b){
    if(!b)return 1;
    if(b&1)return mod_pow(a,b-1)*a%MOD;
    int power=mod_pow(a,b>>1);
    return power*power%MOD;
}

int modular_inverse(int a){
    return mod_pow(a,MOD-2);
}

int lcm(int a, int b){
    return ((a*b)/__gcd(a,b));
}

int nelements(int l, int r){return (abs(r-l)+1);}

int pow2(int a){
    return a*a;
}

int pow3(int a){
    return a*a*a;
}

#define l_shift(x,y) (x<<y)
#define r_shift(x,y) (x>>y)

vector<int> movy={0,0,1,-1};
vector<int> movx={1,-1,0,0};

void solve(){
    map<string,int> fq;
    int n;
    cin>>n;
    string s;
    for(int i=0;i<n;i++){
        cin>>s;
        fq[s]++;
    }
    cout<<"AC x "<<fq["AC"]<<endl;
    cout<<"WA x "<<fq["WA"]<<endl;
    cout<<"TLE x "<<fq["TLE"]<<endl;
    cout<<"RE x "<<fq["RE"]<<endl;
}

int main()
{
    //freopen("IN.txt","r",stdin);
    //freopen("OUT.txt","w",stdout);
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    //int t;cin>>t;while(t--)solve();
    solve();
    return 0;
}