stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include <iostream>

#include <string>

using namespace std;



int main()

{

    long long int n;

    int AC=0, WA=0, TLE=0, RE=0;



    cin>>n;



    string S[n];

    for(int i=0; i<n; i++){

        cin>>S[i];

    }



    for(int j=0; j<n; j++){

        if(S[j] == "AC"){AC++;}

        else if(S[j] == "WA"){WA++;}

        else if(S[j] == "TLE"){TLE++;}

        else if(S[j] == "RE"){RE++;}

    }

    cout<<"AC x "<<AC<<endl;

    cout<<"WA x "<<WA<<endl;

    cout<<"TLE x "<<TLE<<endl;

    cout<<"RE x "<<RE<<endl;



    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

    int t, AC=0, WA=0, TLE=0, RE=0;

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

        if(str[0] == 'A') {

            AC++;

        } else if(str[0] == 'W') {

            WA++;

        } else if(str[0] == 'T') {

            TLE++;

        } else if(str[0] == 'R') {

            RE++;

        }

    }

 

    printf("AC x %d\n", AC);

    printf("WA x %d\n", WA);

    printf("TLE x %d\n", TLE);

    printf("RE x %d\n", RE);

 

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define ll long long

using namespace std;

int main(){

	ll n;

	map<string,ll>m;

	cin >> n;

	for(int i=1; i<=n; i++){

		string s;

		cin >> s;

		m[s]++;

	}

	cout << "AC x " << m["AC"] << endl;

	cout << "WA x " << m["WA"] << endl;

	cout << "TLE x " << m["TLE"] << endl;

	cout << "RE x " << m["RE"] << endl;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

 

int main() {

 

    int t, AC=0, WA=0, TLE=0, RE=0;

    char str[105];

 

    scanf("%d", &t);

    while(t--) {

        scanf("%s", str);

        if(str[0] == 'A') {

            AC++;

        } else if(str[0] == 'W') {

            WA++;

        } else if(str[0] == 'T') {

            TLE++;

        } else if(str[0] == 'R') {

            RE++;

        }

    }

 

    printf("AC x %d\n", AC);

    printf("WA x %d\n", WA);

    printf("TLE x %d\n", TLE);

    printf("RE x %d\n", RE);

 

    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include<bits/stdc++.h>

using namespace std;





void countStr(char ser[], char* arr[],int n)

{

    int coun =0;



    for(int i=0 ; i<n; i++)

    {





        if(strcmp(arr[i],ser) ==0)

        {

            coun++;

        }



    }

    cout<<ser<<" "<<"x"<<" "<<coun<<endl;



}



int main()

{

    int n;

    cin>>n;

    char* arr[n];

    char x[3]="AC";





    if(n>0 && n<10^5)

    {

        cin.ignore();

        for(int i=0 ; i<n; i++)

        {

            arr[i]= new char[4];





            cin.getline(arr[i],4);



            arr[i][4]='\n';

            // cout<<arr[i]<<endl;

            if(strcmp(arr[i],x)!=0   &&  strcmp(arr[i],(char*)"WA")!=0    &&  strcmp(arr[i],(char*)"TLE")!=0   &&  strcmp(arr[i],(char*)"RE")!=0)

            {



                return 0;

            }





        }



        countStr("AC", arr,n);

        countStr("WA", arr,n);

        countStr("TLE",arr,n);

        countStr("RE", arr,n);





    }





    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>



int main() {

	int n, AC = 0, WA = 0, TLE = 0, RE = 0;

	char Si[110];

	

	scanf("%d", &n);

	while(n--){

		scanf("%s", Si);

		if(Si[0] == 'A'){

			AC++;

		}else if(Si[0] == 'W'){

			WA++;

		}else if(Si[0] == 'T'){

			TLE++;

		}else if(Si[0] == 'R'){

			RE++;

	}

}

	printf("AC x %d\n", AC);

	printf("WA x %d\n", WA);

	printf("TLE x %d\n", TLE);

	printf("RE x %d\n", RE);

	

	return 0;



}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define ll long long int

using namespace std;

int main()

{

    int n,i,a=0,w=0,t=0,r=0;

    cin>>n;

    string s[n];

    for(i=0; i<n; i++)

    {

        cin>>s[i];

    }

    for(i=0; i<n; i++)

    {

        if(s[i] == "AC")

            a++;

        else if(s[i] == "TLE")

            t++;

        else if(s[i] == "WA")

            w++;

        else if(s[i] == "RE")

            r++;

    }

    cout<<"AC x "<<a<<endl;

    cout<<"WA x "<<w<<endl;

    cout<<"TLE x "<<t<<endl;

    cout<<"RE x "<<r<<endl;

 

 

 

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int n, ac = 0, wa = 0, tle = 0, re = 0;

	char str[10];

	scanf("%d", &n);

	for(int i = 0; i < n; i++){

		scanf("%s", str);

		if(str[0] == 'A'){

			ac++;

		}

		if(str[0] == 'W'){

			wa++;

		}

		if(str[0] == 'T'){

			tle++;

		}

		if(str[0] == 'R'){

			re++;

		}

	}

	printf("AC x %d\n", ac);

	printf("WA x %d\n", wa);

	printf("TLE x %d\n", tle);

	printf("RE x %d\n", re);

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <string>



using namespace std;



int main()

{

    int arr_size, AC_count = 0, WA_count = 0, TLE_count = 0, RE_count = 0;

    string array[100000], in;

    cin >> arr_size;

    for (int i = 0; i < arr_size; i++)

        cin >> array[i];

    for (int i = 0; i < arr_size; i++)

    {

        if (array[i] == "AC")

            AC_count++;

        else if (array[i] == "WA")

            WA_count++;

        else if (array[i] == "TLE")

            TLE_count++;

        else if (array[i] == "RE")

            RE_count++;

    }

    cout << "AC x " << AC_count << endl << "WA x " << WA_count << endl << "TLE x " << TLE_count << endl << "RE x " << RE_count << endl;

}




// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main()

{

    map<string,int> m;

    int n;

    cin>>n;

    string x;

    for(int i=0;i<n;i++)

    {

        cin>>x;

        m[x]++;

    }

    cout<<"AC"<<" x "<<m["AC"]<<endl;

    cout<<"WA"<<" x "<<m["WA"]<<endl;

    cout<<"TLE"<<" x "<<m["TLE"]<<endl;

    cout<<"RE"<<" x "<<m["RE"]<<endl;

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int main()

{

 int n,a=0,r=0,t=0,w=0;

 cin>>n;

 string s;

 while(n--)

 {

     cin>>s;

     if(s=="AC")a++;

     if(s=="TLE")t++;

     if(s=="RE")r++;

     if(s=="WA")w++;

 }

 cout<<"AC x "<<a<<"\nWA x "<<w<<"\nTLE x "<<t<<"\nRE x "<<r;



    return 0;

}






// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

int N,a,b,c,d;

string x;

int main() {

	cin>>N;

	for(int i=1;i<=N;i++) {

		cin>>x;

		if(x=="AC") {

			a++;

		} else if(x=="WA") {

			b++;

		} else if(x=="TLE") {

			c++;

		} else if(x=="RE") {

			d++;

		}

	}

	cout<<"AC x "<<a<<endl;

	cout<<"WA x "<<b<<endl;

	cout<<"TLE x "<<c<<endl;

	cout<<"RE x "<<d<<endl;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

using namespace std;



int main() {

     int n , arr[4] = {0} ;



     string s , arr_st[4]= { "AC", "WA", "TLE", "RE"} ;



     cin >> n ;



     while(n--){



          cin >> s ;

          for(int i =0 ; i < 4 ; i++){



               if(s == arr_st[i])arr[i]++ ;

          }



     }

     for(int j =0 ; j < 4 ; j++){



          cout << arr_st[j] << " x " << arr[j] << endl ;

     }

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <cmath>

#include <algorithm>

#include <cstdlib>

using namespace std;



int main() {

    int x, a=0,b=0,c=0,d=0;

    scanf("%d",&x);

    string s;

    for (int i = 0; i < x; i++) {

        cin>>s;

        if(s=="AC")

            a++;

        else if (s=="WA")

        b++;

        else if(s=="TLE")

            c++;

        else if(s=="RE")

            d++;

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n",a,b,c,d);



}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



int main()

{

    ios::sync_with_stdio(false);  cin.tie(0); 

    int n , ac=0 ,tle=0, wa=0, re=0;

    string s;

    cin >> n;

    while(n--)

    {

        cin >> s;

        if (s == "AC")

            ac++;

        else if (s == "TLE")

            tle++;

        else if (s == "WA")

            wa++;

        else

            re++;

    }

    cout << "AC x " <<ac <<"\nWA x "<<wa;

    cout << "\nTLE x "<<tle<<"\nRE x "<<re;

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <bits/stdc++.h>



using namespace std;



int main()

{

    int x;

    cin>>x;

    string arrS[x];

    int AC=0, WA=0, TLE=0, RE=0;

    for(int i=0 ; i<x ; i++)

    {

        cin>>arrS[i];

        if(arrS[i]=="AC")

            AC++;

        else if(arrS[i]=="WA")

            WA++;

        else if(arrS[i]=="TLE")

            TLE++;

        else

            RE++;

    }



    cout<<"AC x "<<AC<<endl;

    cout<<"WA x "<<WA<<endl;

    cout<<"TLE x "<<TLE<<endl;

    cout<<"RE x "<<RE;









    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define pi 3.14159

using namespace std;

typedef long long LL;

const LL MOD = 1e9 + 7;

const int N = 1e5 + 7, M = 1e7, OO = 0x3f3f3f3f;

#define AC ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); cin.sync_with_stdio(0);

int main()

{

//have submited with my brother account becuase of forget to swap accounts

int ac=0,wa=0,re=0,tle=0,n;

scanf("%d",&n);

while(n--){

    string s;

    cin>>s;

    if(s=="AC"){

        ac++;

    }

    else if(s=="WA"){

        wa++;

    }

    else if(s=="TLE"){

        tle++;

    }

    else if(s=="RE"){

        re++;

    }

}

printf("AC x %d\n",ac);

printf("WA x %d\n",wa);

printf("TLE x %d\n",tle);

printf("RE x %d\n",re);



    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>



using namespace std;



int main()

{

    int n;

    cin>>n;

    string s;

    int c0=0,c1=0,c2=0,c3=0;

    for (int i=0;i<n;i++)

    {

        cin>>s;

        if (s=="AC")

            c0++;

        else if (s=="WA")

            c1++;

        else if (s=="TLE")

            c2++;

        else if (s=="RE")

            c3++;





    }





    cout<<"AC x "<<c0<<endl;

    cout<<"WA x "<<c1<<endl;

    cout<<"TLE x "<<c2<<endl;

    cout<<"RE x "<<c3;

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

using namespace std;

int n, arr[4];

int main()

{

	cin>>n;

	string str;

	for(int x = 0; x < n; x++){

		cin>>str;

		if(str == "AC") arr[0]++;

		else if(str == "WA") arr[1]++;

		else if(str == "TLE") arr[2]++;

		else if(str == "RE") arr[3]++;



	}



	cout<<"AC x "<<arr[0]<<endl;

	cout<<"WA x "<<arr[1]<<endl;

	cout<<"TLE x "<<arr[2]<<endl;

	cout<<"RE x "<<arr[3]<<endl;



	return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):


#include <bits/stdc++.h>

#include<vector>

using namespace std;

int main()

{

    int N;

    cin>>N;

    int count1=0,count2=0,count3=0,count4=0;

    string s;

    vector<string>v;

    for(int i=0;i<N;i++){

        cin>>s;

        v.push_back(s);

    }

    count1=count(v.begin(),v.end(),"AC");

    count2=count(v.begin(),v.end(),"WA");

    count3=count(v.begin(),v.end(),"TLE");

    count4=count(v.begin(),v.end(),"RE");

    cout<<"AC"<<" "<<"x"<<" "<<count1<<endl;

    cout<<"WA"<<" "<<"x"<<" "<<count2<<endl;

    cout<<"TLE"<<" "<<"x"<<" "<<count3<<endl;

    cout<<"RE"<<" "<<"x"<<" "<<count4<<endl;







    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

#define INF 0x3f3f3f3f

const int N=102,M=10010,mod=998244353;

ll n,m,T,k,cnt,x,y,z,indx,t,ans,a[N];

string s;

int main()

{

	scanf("%lld",&n);

	while (n--)

	{

		cin>>s;

		if (s=="AC") a[1]++;

		if (s=="WA") a[2]++;

		if (s=="TLE") a[3]++;

		if (s=="RE") a[4]++;

	}

	printf("AC x %lld\n",a[1]);

	printf("WA x %lld\n",a[2]);

	printf("TLE x %lld\n",a[3]);

	printf("RE x %lld\n",a[4]);

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>



using namespace std;



int main() {

    int a[4] = {0, 0, 0, 0};

    int i;

    cin >> i;

    while (i--) {

        string s;

        cin >> s;

        if (s == "AC")a[0]++;

        if (s == "WA")a[1]++;

        if (s == "RE")a[2]++;

        if (s == "TLE")a[3]++;

    }

    cout << "AC x " << a[0] << endl;

    cout << "WA x " << a[1] << endl;

    cout << "TLE x " << a[3] << endl;

    cout << "RE x " << a[2] << endl;

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

int i,j,num[4],n;

char s[10];

int main()

{

	scanf("%d",&n);

	for (i=1;i<=n;i++)

	{

		scanf("%s",s);

		if (strcmp(s,"AC")==0) num[0]++;

		if (strcmp(s,"WA")==0) num[1]++;

		if (strcmp(s,"TLE")==0) num[2]++;

		if (strcmp(s,"RE")==0) num[3]++; 

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d",num[0],num[1],num[2],num[3]);

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <iostream>

#include <algorithm>

#include <stack>

using namespace std;



int main(void)

{

  // freopen("aaa.in","r",stdin);

  int N;

  scanf("%d",&N);

  char s[100];

  int numA=0,numW=0,numT=0,numR=0;

  for (int i=1;i<=N;i++)

  {

    scanf("%s",s);

    if (s[0]=='A') numA++;

    if (s[0]=='W') numW++;

    if (s[0]=='T') numT++;

    if (s[0]=='R') numR++;

  }

  printf("AC x %d\n",numA);

  printf("WA x %d\n",numW);

  printf("TLE x %d\n",numT);

  printf("RE x %d",numR);





  return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define ll long long

#define endl "\n"

#define pb push_back

#define READ(FILE) freopen(FILE,"r",stdin)

#define WRITE(FILE) freopen(FILE,"w",stdout)

#define loop(i,s,e) for(int i=s;i<e;i++)

#define mego fast();

using namespace std;

void fast()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

}



ll lcm(ll x, ll y)

{

    return (x*y)/__gcd(x,y);

}

int sum(string s)

{

    int x=0,sz=s.size();

    loop(i,0,sz)x+=(s[i]-'0');

    return x;

}

/*

struct node

{

int data;

node *link;

};

typedef node* nodeptr;

void head_insert(nodeptr& head,int x)

{

    nodeptr tmp;

    tmp=new node;

    tmp->data=x;

    tmp->link=head;

    head=tmp;

}

nodeptr search(nodeptr head,int x)

{

    nodeptr cur=head;

    if(cur ==  nullptr)return cur;

    while(cur->data!=x&&cur->link!=nullptr)

    {

        cur=cur->link;

    }

    if(cur->data==x)return cur;

    else return nullptr;

}

void show(nodeptr head)

{

    nodeptr cur=head;

    if(cur==nullptr)return;

   // cout<<head->data<<" ";

    while(cur->link!=nullptr){

        cout<<cur->data<<" ";

        cur=cur->link;

    }

    cout<<endl;

}

*/



int main()

{

    mego

    int n,a=0,b=0,c=0,d=0;

    string s;

    cin>>n;

    loop(i,0,n)

    {

        cin>>s;

        if(s=="AC")

            a++;

        else if(s=="WA")

            b++;

        else if(s=="TLE")

            c++;

        else

            d++;



    }

    cout<<"AC x "<<a<<endl;

    cout<<"WA x "<<b<<endl;

    cout<<"TLE x "<<c<<endl;

    cout<<"RE x "<<d;

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);

    int n, ac = 0, wa = 0, tle = 0, re = 0; cin >> n;

    string s;

    while (n--)

    {

        cin >> s;

        if(s == "AC") ac++;

        else if(s == "WA") wa++;

        else if(s == "TLE") tle++;

        else re++;

    }

    cout << "AC x " << ac << "\nWA x " << wa << "\nTLE x " << tle << "\nRE x " << re << '\n';

    return 0;

}

// Optimized solution to the same problem (faster):
#include<stdio.h>



int main(){

	int N;

	int c1=0,c2=0,c3=0,c4=0;

	scanf("%d", &N);

	for(int i=0; i<N; i++){

		char msg[5];

		scanf("%s", msg);

		if(msg[0]=='A'){

			c1++;

		}

		else if(msg[0]=='W'){

			c2++;

		}

		else if(msg[0]=='T'){

			c3++;

		}

		else{

			c4++;

		}

	}

	printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n", c1, c2, c3, c4);

	return 0;

}

// Fastest known solution for this problem: