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;

#define fastio cin.tie(NULL); cout.tie(NULL); ios_base::sync_with_stdio(0)
typedef long long ll;
//#define int ll
#define swap(x, y)((x)^=(y)^=(x)^=(y))
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vector<int>> matrix;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define all(c) c.begin(),c.end()
#define mp make_pair
#define pb push_back
#define f first
#define tcase int _; cin>> _; for( ; _; --_)
#define s second
#define endl '\n'
const int MAX = 1e6+1;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1);
const double DEG = 180/PI;
const int MOD = 1e9+7;
int mod(int x, int m){ if(x >= 0) return x % m; if(-x < m) return m-(-x); return mod(x%m, m);}
int gcd (int a, int b) {while(b){a = mod(a, b); swap(a, b);} return a;}
int lcm (int a, int b) {return a / gcd(a, b) * b;}
int fexp(int base, int exp, int m){
	int total = 1;
	while(exp > 0){
		if(exp % 2 == 1){
			total = mod(((mod(total, m)) * (mod(base, m))), m);
			exp--;
		}
		base= mod((mod(base, m) * mod(base, m)), m);
		exp /= 2;
	}
	return mod(total, m);
}

int n;

void solve(){
    cin>> n;
    vi r(4, 0);
    vector<string> st = {"AC", "WA", "TLE", "RE"};
    for(int i = 0; i <n; ++i){
        string s; cin>> s;
        if(s == "AC") r[0]++;
        else if(s == "WA") r[1]++;
        else if(s == "TLE") r[2]++;
        else r[3]++;
    }
    for(int i= 0; i<4; ++i) cout<< st[i] << " x "<< r[i]<< endl;
}

int32_t main(){
	fastio;
	solve();
	return 0;
}