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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.NoSuchElementException;
public class Main {
	public static void main(String[] args)throws Exception{
		PrintWriter out = new PrintWriter(System.out);
		int N=nextInt();
		int A=0,B=0,C=0,D=0;
		for(int i=0;i<N;i++){
			String a=next();
			if(a.equals("AC"))
				A++;
			if(a.equals("WA"))
				B++;
			if(a.equals("TLE"))
				C++;
			if(a.equals("RE"))
				D++;
		}
		System.out.println("AC x "+A);
		System.out.println("WA x "+B);
		System.out.println("TLE x "+C);
		System.out.println("RE x "+D);
	}
	static boolean isOK(int index,int key,ArrayList<Integer>a){
		if (a.get(index) >= key) return true;
	    else return false;
	}
	static int binary_search(int key,ArrayList<Integer> a) {
	    int ng = -1; //「index = 0」が条件を満たすこともあるので、初期値は -1
	    int ok = (int)a.size(); // 「index = a.size()-1」が条件を満たさないこともあるので、初期値は a.size()
	    while (Math.abs(ok - ng) > 1) {
	        int mid = (ok + ng) / 2;

	        if (isOK(mid,key,a)) ok = mid;
	        else ng = mid;
	    }
	    return ok;
	}
	//FastScanner
	static InputStream in = System.in;
	static  byte[] buffer = new byte[1024];
	static int length = 0, p = 0;
	public static boolean hasNextByte () {
		if (p < length) return true;
		else {
			p = 0;
			try {length = in.read(buffer);}
			catch (Exception e) {e.printStackTrace();}
			if (length == 0) return false;
		}
		return true;
	}
	public static int readByte () {
		if (hasNextByte() == true) return buffer[p++];
		return -1;
	}
	public static boolean isPrintable (int n) {return 33<=n&&n<=126;}
	public static void skip () {
		while (hasNextByte() && !isPrintable(buffer[p])) p++;
	}
	public static boolean hasNext () {skip(); return hasNextByte();}
	public static String next () {
		if (!hasNext()) throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int temp = readByte();
		while (isPrintable(temp)) {
			sb.appendCodePoint(temp);
			temp = readByte();
		}
		return sb.toString();
	}
	public static int nextInt () {return Math.toIntExact(nextLong());}
	public static int[] nextInts (int n) {
		int[] ar = new int[n];
		for (int i=0; i<n; i++) ar[i] = nextInt();
		return ar;
	}
	public static long nextLong () {
		if (!hasNext()) throw new NoSuchElementException();
		boolean minus = false;
		int temp = readByte();
		if (temp == '-') {
			minus = true;
			temp = readByte();
		}
		if (temp<'0' || '9'<temp) throw new NumberFormatException();
		long n = 0;
		while (isPrintable(temp)) {
			if ('0'<=temp && temp<='9') {
				n *= 10;
				n += temp - '0';
			}
			temp = readByte();
		}
		return minus? -n : n;
	}
}