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

#include <cstdio>

#include <algorithm>

#include <vector>

#include <cmath>

#include <cstring>

#define ll long long

using namespace std;



int n,num[10];

char s[10];



int main() {

    scanf("%d",&n);

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

        scanf("%s",s);

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

        if (s[0]=='W') ++num[1];

        if (s[0]=='T') ++num[2];

        if (s[0]=='R') ++num[3];

    }

    printf("AC x %d\n",num[0]);

    printf("WA x %d\n",num[1]);

    printf("TLE x %d\n",num[2]);

    printf("RE x %d\n",num[3]);

    return 0;

}


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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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

using namespace std;

int main()

{

    int n;

    cin>>n;

    int a,b,c,d;

    a=b=c=d=0;

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

    {

        string s;

        cin>>s;

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

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

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

        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 c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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

#define ll long long

#define _ ios::sync_with_stdio(false);cout.tie(NULL);cin.tie(NULL);

using namespace std;



int main() { _

    int n; cin >> n;

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

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

        string s; cin >> s;

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

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

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

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

    }

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

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

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

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



}

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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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

*         ****FOX****

*/

#include <iostream>

#include <string>

#include <algorithm>

#include <stack>

#include <vector>

#include <map>

#include <set>

#include <sstream>

#include <iomanip>

#include <cmath>

#include <limits>

//freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);

#include <utility>

#include <list>

#include <cstdio>

#include<forward_list>

#include<algorithm>

#include<string>

#define Omar_Hafez ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define pi 3.1415926536

#define ll long long

#define ull unsigned long long

#define lli long long int

using namespace std;

int main()

{

    Omar_Hafez

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

    string x;

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

    {

        cin >> x; 

        if (x[0] == 'A') a++;

        else if (x[0] == 'W') w++;

        else if (x[0] == 'T')t++;

        else if (x[0] == 'R')r++;

    }

    cout << "AC " << "x " << a << "\n";

    cout << "WA " << "x " << w << "\n";

    cout << "TLE " << "x " <<t << "\n";

    cout << "RE " << "x " << r << "\n";

        return 0;

}


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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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

#include<bits/stdc++.h>

using namespace std;



int main()

{

      int n;

      cin>>n;

      int ac=0;

      int wa=0;

      int tle=0;

      int re=0;

      vector<string> s(n);

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

        cin>>s[i];

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

          ++ac;

        }

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

          ++wa;

        }

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

          ++tle;

        }

        else if(s[i]=="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 c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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

using namespace std;



int main() {

  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);

  int n;

  cin >> n;

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

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

    string in;

    cin >> in;

    if (in == "AC") AC++;

    if (in == "TLE") TLE++;

    if (in == "WA") WA++; 

    if (in == "RE") RE++;

  }

  cout << "AC x " << AC << '\n' << "WA x " << WA << '\n' << "TLE x " << TLE << '\n' << "RE x " << RE << '\n';   

}            


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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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



using namespace std;



int main()

{

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

    cin >> n ;

    string arr[n] ;

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

        cin>> arr[i] ;

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

           c0=c0+1 ;

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

            c1=c1+1 ;

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

            c2=c2+1 ;

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

            c3=c3+1 ;



    }

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

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

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

    cout << "RE x "<< c3 << endl ;

    return 0;

}


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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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



using namespace std;



void fast();



/*

*

* Watch out! Another Buggy Code From Muhammad_Mamdouh, Happy Coding;

*

*/



int main()

{

    fast();

    int n; cin >> n;

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



    while (n--)

    {

        string s; cin >> s;



        if (s == "AC")

            ++a;

        if (s == "WA")

            ++b;

        if (s == "TLE")

            ++c;

        if (s == "RE")

            ++d;

    }



    cout << "AC" << " x " << a << '\n';

    cout << "WA" << " x " << b << '\n';

    cout << "TLE" << " x " << c << '\n';

    cout << "RE" << " x " << d << '\n';



    return 0;

}



void fast()

{

    ios_base::sync_with_stdio(0);

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

}

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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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

using namespace std;

int main() {

    int countA = 0,countW = 0, countT = 0, countR = 0;

    int n;

    cin>>n;

    string s [n];

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

        cin >> s[i];

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

            countA++;

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

            countW++;

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

            countT++;

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

            countR++;

    }

    cout<<"AC x "<<countA<<"\n";

    cout<<"WA x "<<countW<<"\n";

    cout<<"TLE x "<<countT<<"\n";

    cout<<"RE x "<<countR<<"\n";

    return 0;

}

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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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

using namespace std;

int main (){

    int n;cin>>n;string a[n];int cnt1=0,cnt2=0,cnt3=0,cnt4=0;

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

            cin>>a[i];

            if (a[i]=="AC")cnt1++;

            else if (a[i]=="WA")cnt2++;

            else if (a[i]=="TLE")cnt3++;

            else {cnt4++;}



    }

    cout<<"AC"<<" "<<"x"<<" "<<cnt1<<endl;

    cout<<"WA"<<" "<<"x"<<" "<<cnt2<<endl;

    cout<<"TLE"<<" "<<"x"<<" "<<cnt3<<endl;

    cout<<"RE" <<" "<<"x"<<" "<<cnt4<<endl;

    return 0;

}


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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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



using namespace std;



int main()

{

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

    char c;



    cin >> n;

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

    {

        cin >> c;

        switch (c)

        {

        case 'A':

            ++ac;

            cin.ignore(1);

            break;

        case 'W':

            ++wa;

            cin.ignore(1);

            break;

        case 'T':

            ++tle;

            cin.ignore(2);

            break;

        case 'R':

            ++re;

            cin.ignore(1);

            break;

        }

    }



    cout << "AC" << " x " << ac << '\n';

    cout << "WA" << " x " << wa << '\n';

    cout << "TLE" << " x " << tle << '\n';

    cout << "RE" << " x " << re << '\n';

    return 0;

}

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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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



using namespace std;



int main()

{

int n,A=0,B=0,C=0,D=0;

cin>>n;

string s;

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

    cin>>s;

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

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

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

    else if (s=="RE")D++;

}

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

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

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

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

    return 0;

}


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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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

#include <string>

using namespace std;

int main() {



	long long n;

	string x;

	cin >> n;



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



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

	{

		cin >> x;

		

		if (x == "AC")

			AC++;

		else if (x == "WA")

			WA++;

		else if (x == "TLE")

			TLE++;

		else if (x == "RE")

			RE++;





	}



	cout << "AC x " << AC;

	cout << endl;

	cout << "WA x " << WA;

	cout << endl;

	cout << "TLE x " << TLE;

	cout << endl;

	cout << "RE x " << RE;

}

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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  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[n];

    int a,w,t,r;

    a=w=t=r=0;

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

        cin>>s[i];

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

           a++;

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

            w++;

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

            t++;

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

            r++;

    }



    cout << "AC x " <<a<<"\n";

    cout << "WA x " <<w<<"\n";

    cout << "TLE x " <<t<<"\n";

    cout << "RE x " <<r<<"\n";

    return 0;

}


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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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

#include <map>



using namespace std;



int main()

{

    int n;

    cin>>n;

    map<string,int> counter;



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

    {

        string temp;

        cin>>temp;

        counter[temp]++;

    }



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

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

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

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



    return 0;

}


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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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

#include<string>

#include<algorithm>

#include<vector>

using namespace std;

int main()

{

	int n;

	cin >> n;

	int a, s, d, f;

	a = s = d = f = 0;

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

	{

		string x;

		cin >> x;

		if (x=="AC")

			a++;

		if (x == "WA")

			s++;

		if (x == "TLE")

			d++;

		if (x == "RE")

			f++;

	}

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

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



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

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



	return 0;

}

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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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



using namespace std;



int main()

{

	int n, C1 = 0, C2 = 0, C3 = 0, C4 = 0;

	cin>>n;

	string o;

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

     {

          cin>>o;

          if ( o == "AC" )

               C1 += 1;

          else if ( o == "WA" )

               C2 += 1;

          else if ( o == "TLE" )

               C3 += 1;

          else if ( o == "RE" )

               C4 += 1;

     }

     cout<<"AC x "<<C1<<endl;

     cout<<"WA x "<<C2<<endl;

     cout<<"TLE x "<<C3<<endl;

     cout<<"RE x "<<C4<<endl;

    return 0;

}


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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

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

using namespace std;



int a;



int main() {

	map<string, int> m;

	cin>>a;

	for(int k = 1; k<= a; k++){

		string s;

		cin >> s;

		m[s] ++;

	}

	cout << "AC x " << m["AC"]<< '\n';

	cout << "WA x " << m["WA"] << '\n';

	cout << "TLE x " << m["TLE"] << '\n';

	cout << "RE x " << m["RE"]<< '\n';

}

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



int main() {

  int c, verdict[4] = {0}; 

  scanf("%d", &c);



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

    char str[5];

    scanf("%s", str);

    switch(str[0]) {

      case 'A':

        verdict[0]++;

        break;

      case 'W':

        verdict[1]++;

        break;

      case 'T':

        verdict[2]++;

        break;

      case 'R':

        verdict[3]++;

        break;

    }

  }



  printf("AC x %d\n", verdict[0]);

  printf("WA x %d\n", verdict[1]);

  printf("TLE x %d\n", verdict[2]);

  printf("RE x %d\n", verdict[3]);

  return 0;

}

// Fastest known solution for this problem: