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

#include <string>

#include <bits/stdc++.h>

#include <vector>

using namespace std;



int main()

{

    int t, ac = 0, wa = 0, tl = 0, re = 0;

    string s;

    cin >> t;

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

    {



        cin >> s;

        if (s == "AC")

            ac++;

        else if (s == "WA")

            wa++;

        else if (s == "TLE")

            tl++;

        else

            re++;

    }

    cout << "AC x " << ac << "\n" << "WA x " << wa << "\n" << "TLE x " << tl << "\n" << "RE x " << re;

    return 0;

}


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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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

#include <iostream>

#include <cstring>

#include <set>

#include <algorithm>

#include <math.h>

#include <vector>

#include <iterator>

#include<deque>

#include<cstdio>



#define lp(i,a,n) for(int i=a;i<n;++i)

#define rep(i,a,n) for(ll i=a;i<=n;++i)

#define all(v) v.begin(),v.end()

#define clr(n,a) memset(n,a,sizeof(n))

#define vi vector <int>

#define sc set <char>

#define si set<int>

#define pb push_back

#define vb vector <bool>

#define pi(n) printf("%d\n",n)

#define ln "\n"

#define mahmoud_abdelsamie ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

typedef long long ll;

typedef unsigned long long ull;

using namespace std;





int main()

{

    mahmoud_abdelsamie

    int a[4]={};

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

    int n;cin>>n;

    lp(i,0,n){

        string st; cin>>st;

        if(st==s[0])

            ++a[0];

        else if(st==s[1])

            ++a[1];

        else if(st==s[2])

            ++a[2];

        else

            ++a[3];

    }

    lp(i,0,4){

        cout<<s[i] << " x "<<a[i]<<ln;

    }

    return 0;

}


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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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

// Muhammad Eid //

#include <bits/stdc++.h>

using namespace std;



#define ll long long

#define ull unsigned long long

#define prINT pair<int, int>



const int N = 4 +5;

int ar[N];



void Mo35(){ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);}



int main(){

Mo35();



    string s;

    int t; cin>> t;

    while(t--){

        cin>> s;

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

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

        else if(s == "TLE") ar[2]++;

        else ar[3]++;

    }

    cout<< "AC x "<< ar[0]<< "\n";

    cout<< "WA x "<< ar[1]<< "\n";

    cout<< "TLE x "<< ar[2]<< "\n";

    cout<< "RE x "<< ar[3]<< "\n";



return 0;

}


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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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

#include <vector>

using namespace std;

void creatVec(vector<string> &vec, size_t size, size_t &AC, size_t &WA, size_t &TLE, size_t &RE);

void check(const string &s, size_t &AC, size_t &WA, size_t &TLE, size_t &RE);

void printer(const string &s, size_t n);

int main() {

    size_t n{}, countAC{}, countWA{}, countTLE{}, countRE{};

    cin >> n;

    vector<string> words {};

    creatVec(words, n, countAC, countWA, countTLE, countRE);

    printer("AC", countAC);

    printer("WA", countWA);

    printer("TLE", countTLE);

    printer("RE", countRE);

    return 0;

}

void creatVec(vector<string> &vec, size_t size, size_t &AC, size_t &WA, size_t &TLE, size_t &RE) {

    for (size_t i{}; i < size; i++) {

        string in{};

        cin >> in;

        check(in, AC, WA, TLE, RE);

        vec.push_back(in);

    }

}

void check(const string &s, size_t &AC, size_t &WA, size_t &TLE, size_t &RE) {

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

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

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

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

}

void printer(const string &s, size_t n) {cout << s << " x " << n << endl;}

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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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



using namespace std;







int main() {



    int n;

    cin>>n;

    string word;

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

    while(n--){

        cin>>word;

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

            ac++;

        }

        else if(word[0]=='R'){

            re++;

        }

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

            tle++;

        }

        else {

            wa++;

        }



    }



    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() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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

using namespace std;

int n;

string s;

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

int main()

{

	cin>>n;

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

	{

		cin>>s;

		if(s=="AC") c1++;

		else if(s=="WA")  c2++;

		else if(s=="TLE") c3++;

		else              c4++;

	}

	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() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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

#define ll long long

using namespace std;



const char* s[5]={"","AC","WA","TLE","RE"};

int a[10],n;

char st[10];



int main() {

	scanf("%d",&n);

	a[1]=a[2]=a[3]=a[4]=0;

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

		scanf("%s",st);

		if (st[0]=='A') a[1]++;

		if (st[0]=='W') a[2]++;

		if (st[0]=='T') a[3]++;

		if (st[0]=='R') a[4]++;

	}

	for (int i=1;i<=4;++i) {

		printf("%s x %d\n",s[i],a[i]);

	}

}

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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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

#include <cmath>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <string>

using namespace std;

typedef long long ll;



const int maxn = 1e5 + 5;

int c[5];



int main() {

    int n;

    cin>>n;

    string s;

    while(n-->0){

        cin>>s;

        if(s[0]=='A') c[1]++;

        else if(s[0]=='W') c[2]++;

        else if(s[0]=='T') c[3]++;

        else if(s[0]=='R') c[4]++;

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d\n",c[1],c[2],c[3],c[4]);

    return 0;

}

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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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

//  main.cpp

//  B - Judge Status Summary

//

//  Created by 李远铄 on 2020/8/7.

//  Copyright © 2020 李远铄. All rights reserved.

//



#include <iostream>

#include <string>



#define _DEBUG_

#ifdef _DEBUG_

#define debug(x) cout << #x << ": " << x << "  "

#define print_n printf("\n")

#define cout_n cout<<'\n'

#else

#define debug(x)

#define print_n

#define cout_n

#endif





using namespace std;

char s[10];

int a,w,t,r;

int main(int argc, const char * argv[]) {

    int n;

    scanf("%d",&n);

    while(n--){

        scanf("%s",s);

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

            a++;

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

            w++;

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

            t++;

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

            r++;

        }

    }

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

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

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

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

    return 0;

}


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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
						/* In the name of Allah */

#include<bits/stdc++.h>

using namespace std;

int main()

{

	int t;

	cin >> t;

	map<string, int>m;

	while (t--)

	{

		string x;

		cin >> x;

		m[x]++;

	}

	cout << "AC x " << m["AC"] << endl << "WA x " << m["WA"] << endl << "TLE x " << m["TLE"] << endl << "RE x " << m["RE"] << endl;

	return 0;

}










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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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



using namespace std;



int main() {

    cin.tie(0);

    cout.tie(0);

    ios_base::sync_with_stdio(0);



    int n;

    cin >> n;

    map<string, int> mp;

    string s;



    while (n--) {

        cin >> s;

        ++mp[s];

    }

    vector<string> v = {"AC", "WA", "TLE", "RE"};

    for (auto i : v) {

        cout << i << " x " << mp[i] << '\n';

    }





}

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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

// 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,t=0;

    cin>>n;

    string s;

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

    {

        cin>>s;

        if(s=="AC")

        {

            a++;

        }

        if(s=="TLE")

        {

            t++;

        }

        if(s=="WA")

        {

            b++;

        }

        if(s=="RE")

        {

            c++;

        }

    }

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

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

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

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

    return 0;

}


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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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



using namespace std;



int main()

{

string x;



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



cin>>n;

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

cin>>x;

    if(x=="AC"){AC++;}

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

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

else if(x=="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() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

// 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;

    cin >> N;

    string entr;

    if (N<=0 || N>100000){return 0;}

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

        cin >> entr;

        if (entr == "AC") {

            AC ++;

        }

        if (entr == "WA"){

            WA++;

        }

        if (entr == "TLE"){

            TLE++;

        }

        if (entr == "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() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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

#include<vector>

using namespace std;

int main()

{

	int n;

	cin >> n;

	string s;

	vector<pair<string, int>>v;

	v.push_back(make_pair("AC", 0));

	v.push_back(make_pair("WA", 0));

	v.push_back(make_pair("TLE", 0));

	v.push_back(make_pair("RE", 0));

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

	{

		cin >> s;

		if (s == "AC")

			v[0].second++;

		else if (s == "WA")

			v[1].second++;

		else if (s == "TLE")

			v[2].second++;

		else if (s == "RE")

			v[3].second++;

	}

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

	{

		cout << v[i].first << " " << "x" << " " << v[i].second;

		if (i < 3)

		{

			cout << endl;

		}

	}

	return 0;

}

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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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


#include <bits/stdc++.h>

using namespace std;

#include<vector>

 int main()

   {

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

    int n;

    cin>>n;

    string s;

    while(n--)

      {

        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() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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

#include <algorithm>

#include <string>

#define ll long long

using namespace std;

int main(){

    string s;

    ll n;

    string ac="AC";

    ll cac=0;

    string wa="WA";

    ll cwa=0;

    string tle="TLE";

    ll ctle=0;

    string re="RE";

    ll cre=0;

    cin>>n;

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

        cin>>s;

        if(s==ac){

            cac++;

        }

        else if(s==wa){

            cwa++;

        }

        else if(s==tle){

            ctle++;

        }

        else if (s==re){

            cre++;

        }

    }

    cout<<ac<<" x "<<cac<<endl;

    cout<<wa<<" x "<<cwa<<endl;

    cout<<tle<<" x "<<ctle<<endl;

    cout<<re<<" x "<<cre<<endl;

}

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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

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

#include <vector>

#include <algorithm>



using namespace std;



int main()

{

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

    string s;



    cin >> n;

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

    {

        cin >> s;

        if (s == "AC")

            AC++;

        else if (s == "WA")

            WA++;

        else if (s == "TLE")

            TLE++;

        else if (s == "RE")

            RE++;

    }

    cout << "AC x " << AC << "\nWA x " << WA << "\nTLE x " << TLE << "\nRE x " << RE << endl;

    return 0;

}


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



int main() {

    long int t, ac = 0, wa = 0, tle = 0, re = 0;

    scanf("%ld", &t);

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

        char temp[4];

        scanf("%s", &temp);

        switch (temp[0]) {

            case 'A':

                ac++;

                break;

            case 'W':

                wa++;

                break;

            case 'T':

                tle++;

                break;

            case 'R':

                re++;

                break;

        }

    }

    printf("AC x %d\nWA x %d\nTLE x %d\nRE x %d", ac, wa, tle, re);

}

// Fastest known solution for this problem: