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 <iostream>
#include <vector>
#include <deque>
#include <algorithm>
#include <numeric>
#include <string>
#include <cstring>
#include <list>
#include <unordered_set>
#include <tuple>
#include <cmath>
#include <limits>
#include <type_traits>
#include <iomanip>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <set>
#include <bitset>
#include <regex>
#include <random>
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
#define rep(i,n)for(ll i=0;i<n;++i)
#define exout(x) printf("%.10f\n", x)
const double pi = acos(-1.0);
const ll MOD = 1000000007;
const ll INF = 1e10;
const ll MAX_N = 1010;

//組み合わせの余りを求める
ll fac[MAX_N], finv[MAX_N], inv[MAX_N];

void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX_N; i++) {
		fac[i] = fac[i - 1] * i;
	}
}

// 二項係数計算
long long COM(ll n, ll k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	if (n > k * 2) {

	}
	return fac[n] / (fac[k] * fac[n - k]);
}

ll dp[201010][2];
ll dx[4] = { 0,0,-1,1 };
ll dy[4] = { -1,1,0,0 };

//long longしか使わない
//素数は1より大きい
int main() {
	ll n;
	cin >> n;
	vector<string>s(n);
	vector<ll>ans(4);
	vector<string>res(4);
	res[0] = "AC";
	res[1] = "WA";
	res[2] = "TLE";
	res[3] = "RE";
	rep(i, n) {
		cin >> s[i];
		if (s[i] == "AC")ans[0]++;
		else if (s[i] == "WA")ans[1]++;
		else if (s[i] == "TLE")ans[2]++;
		else {
			ans[3]++;
		}
	}

	rep(i, 4) {
		cout << res[i] << " " << "x" << " " << ans[i] << endl;
	}
	return 0;
}