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>
using namespace std;

typedef long long ll;
typedef long double ld;

#define pb push_back
#define sort(s) sort(s.begin(),s.end())
#define reverse(s) reverse(s.begin(),s.end())
#define rep(i,n) for(ll (i)=0;(i)<(n);(i)++)

const ll mod = 1e9+7;

//最大公約数
ll gcd(ll a, ll b) {
    return b ? gcd(b, a%b) : a;
}

//最小公倍数
ll lcm(ll a, ll b) {
    return a / gcd(a, b) * b;
}

//素数判定
bool isPrime(ll x){
    ll i;
    if(x < 2)return 0;
    else if(x == 2) return 1;
    if(x%2 == 0) return 0;
    for(i = 3; i*i <= x; i += 2) if(x%i == 0) return 0;
    return 1;
}

//桁和
int digsum(ll n) {
    int res = 0;
    while(n > 0) {
        res += n%10;
        n /= 10;
    }
    return res;
}

//桁数
int dignum(ll n) {
    int res = 0;
    while(n > 0) {
        res++;
        n /= 10;
    }
    return res;
}

//文字列中の特定の文字カウント
ll stringcount(string s,char c){
    return count(s.cbegin(),s.cend(),c);
}

//階乗
ll ka(ll i){
    ll res=1;
    while(i>0){
        res=res*i;
        i--;
    }
    return res;
}
 
//ncr
ll ncr(ll x,ll y){
    ll a = x;
    ll b = 1;
    if(y>(x/2)){
        y=x-y;
    }
    for(ll i=1;i<y;i++){
        a*=x-i;
        b*=i+1;
    }
    return a/b;
}

//フィボナッチ数列
ll f(ll x){
    ll dp[x+1];
    dp[1]=1;
    dp[2]=1;
    for(ll i=3;i<=x;i++){
        dp[i]=dp[i-1]+dp[i-2];
    }
    return dp[x];
}

//char->int
int ctoi(char c){
    return c-'0';
}

vector<pair<long long, long long> > prime_factorize(long long N) {
    vector<pair<long long, long long> > res;
    for (long long a = 2; a * a <= N; ++a) {
        if (N % a != 0) continue;
        long long ex = 0; // 指数

        // 割れる限り割り続ける
        while (N % a == 0) {
            ++ex;
            N /= a;
        }

        // その結果を push
        res.push_back({a, ex});
    }

    // 最後に残った数について
    if (N != 1) res.push_back({N, 1});
    return res;
}



int main(){
    int n;
    cin >> n;
    string s[n];
    rep(i,n) cin >> s[i];

    map<string,int> mp;

    for(int i=0;i<n;i++){
        mp[s[i]]++;
    }

    cout << "AC x " << mp["AC"] << endl;
    cout << "WA x " << mp["WA"] << endl;
    cout << "TLE x " << mp["TLE"] << endl;
    cout << "RE x " << mp["RE"] << endl;
}