stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
 #include <stdio.h>

#include <math.h>

#include <string.h>

#include <algorithm>

#include <iostream>

#include <string>

#include <time.h>

#include <queue>

#include <list>

#include <string.h>

#define sf scanf

#define pf printf

#define lf double

#define ll long long

#define p123 printf("123\n");

#define pn printf("\n");

#define pk printf(" ");

#define p(n) printf("%d",n);

#define pln(n) printf("%d\n",n);

#define s(n) scanf("%d",&n);

#define ss(n) scanf("%s",n);

#define ps(n) printf("%s",n);

#define sld(n) scanf("%lld",&n);

#define pld(n) printf("%lld",n);

#define slf(n) scanf("%lf",&n);

#define plf(n) printf("%lf",n);

#define sc(n) scanf("%c",&n);

#define pc(n) printf("%c",n);

#define gc getchar();

#define re(n,a) memset(n,a,sizeof(n));

#define len(a) strlen(a)

#define LL long long

#define eps 1e-6

using namespace std;

const double pi = 3.1415926535;

/*

https://codeforces.com/contest/1106/problems

https://codeforces.com/contest/1106/submit

*/

const ll mod = 1e9+7;

//快速幂 (a^b)%mod

//如果求逆元，则b = mod-2;

ll pow_quick(ll a,ll b){

    ll r = 1,base = a;

    while(b){

        if(b & 1){

            r *= base;

            r %= mod;

        }

        base *= base;

        base %= mod;

        b >>= 1;

    }

    return r;

}



char a[1000000];

int main(){

    int h,w;

    s(h) s(w)

    gc

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

        ss(a) gc

        puts(a);

        puts(a);

    }

	return 0;

 }/*

int t[100000];

int main(){

    int n = 0;

    s(n)

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

        s(t[i])

    }

    int m;

    s(m)





	return 0;

 }

*/


// Faster C++ solution for the same problem (measured speedup: 5.066772474694874x):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

struct chu

{

    char t[101];

}s[101];

int main()

{

      int a,b;

      while(~scanf("%d%d",&a,&b))

      {

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

         {

                scanf("%s",s[i].t);

         }

         for(int j=1;j<=a;j++)

         {

             printf("%s\n",s[j].t);

             printf("%s\n",s[j].t);

         }

      }

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

#define ll long long

#define pi acos(-1,0)

int main()

{

   int m,n;

   cin>>m>>n;

   string s;

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

   {

       cin>>s;

       cout<<s<<endl;

       cout<<s<<endl;

   }

   return 0;

   // 2 2 4 1 3

   //

}


// Faster C++ solution for the same problem (measured speedup: 5.216705528946235x):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

struct chu

{

    char t[101];

}s[101];

int main()

{

      int a,b;

      while(~scanf("%d%d",&a,&b))

      {

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

         {

                scanf("%s",s[i].t);

         }

         for(int j=1;j<=a;j++)

         {

             printf("%s\n",s[j].t);

             printf("%s\n",s[j].t);

         }

      }

    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;



int main() {

    int h,w;

    string s;

    scanf("%d %d",&h,&w);

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

        cin>>s;

        cout<<s<<'\n'<<s<<'\n';

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.173499495535133x):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

struct chu

{

    char t[101];

}s[101];

int main()

{

      int a,b;

      while(~scanf("%d%d",&a,&b))

      {

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

         {

                scanf("%s",s[i].t);

         }

         for(int j=1;j<=a;j++)

         {

             printf("%s\n",s[j].t);

             printf("%s\n",s[j].t);

         }

      }

    return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;



int main() {

	int x, y;

	cin >> x >> y;

	string str, temp[1001];

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

		cin >> str;

		temp[i] = str;

	}

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

		cout << temp[i] << endl;

		cout << temp[i] << endl;

	}



}

// Faster C++ solution for the same problem (measured speedup: 5.366658135939175x):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

struct chu

{

    char t[101];

}s[101];

int main()

{

      int a,b;

      while(~scanf("%d%d",&a,&b))

      {

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

         {

                scanf("%s",s[i].t);

         }

         for(int j=1;j<=a;j++)

         {

             printf("%s\n",s[j].t);

             printf("%s\n",s[j].t);

         }

      }

    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main()

{

    int a,b,i;

    cin>>a>>b;

    string s[a];

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

        cin>>s[i];

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

        {

            cout<<s[i]<<endl<<s[i]<<endl;

        }



    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 5.219146948470865x):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

struct chu

{

    char t[101];

}s[101];

int main()

{

      int a,b;

      while(~scanf("%d%d",&a,&b))

      {

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

         {

                scanf("%s",s[i].t);

         }

         for(int j=1;j<=a;j++)

         {

             printf("%s\n",s[j].t);

             printf("%s\n",s[j].t);

         }

      }

    return 0;

}
// Slower C++ solution:
#include <iostream>

#include <cmath>

using namespace std;



int main() {

    string arr[100],arr2[200][100];

    int h,w; cin>>h>>w;

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

        cin>>arr[i];

    }

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

        cout<<arr[i]<<endl;

        cout<<arr[i]<<endl;

    }

}

// Faster C++ solution for the same problem (measured speedup: 1.6652962886656952x):
                #include <bits/stdc++.h>

                using namespace std;

                typedef long long ll;

                #define INPUT()              freopen("input.txt", "r", stdin)

                #define OUTPUT()             freopen("output.txt", "w", stdout)

                #define F                    first

                #define S                    second

            //Templatefunctions

                template<typename T>T gcd(T x, T y) { if (y == 0)return x; else return gcd(y, x%y);}

                template <class T> T euclideanDistance(T x1 , T x2 , T y1 , T y2){return sqrt(( x1 - x2 ) * ( x1 - x2 ) + ( y1 - y2 ) * ( y1 - y2 ));}

                template <class T> T digitSum(T n){T sum = 0;while(n > 0)sum += n % 10,n /= 10;return sum;}

            //datatypes

                typedef long long int ll;

                typedef unsigned long long int ull;

                typedef long double ld;

            //constants

                const double INF = 1e+9;

                const double INFLL = 1e+18;

                const double EPS = 1e-9;

                const double PI = acos(-1);

                const int knightDir[8][2] = {{-2,-1},{-2,1},{-1,2},{1,2},{2,-1},{2,1},{-1,-2},{1,-2}};

                       //const int dx[]={0,0,0,1,1,1,-1,-1,-1};

                       //const int dy[]={-1,0,1,-1,0,1,-1,0,1};

                const int dx[] = {0 , 1 , 0 , -1};

                const int dy[] = {1 , 0 , -1 , 0};

            //helpers

                //#define MOD         1000000007LL

                #define MAX         1000005

                const ll MOD = 998244353;

                int arr[MAX];

            //Global Variables







                int main()

                {

                        int x , y;

                        scanf("%d%d",&x,&y);



                        string arr[x];



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

                        {

                            cin>>arr[i];

                            //cout<<arr[i]<<"\n"<<arr[i]<<"\n";

                        }

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

                        {

                            //cin>>arr[i];

                            cout<<arr[i]<<"\n"<<arr[i]<<"\n";

                        }

                }
// Slower C++ solution:
#include <iostream>

#include <cmath>

using namespace std;



int main() {

    string arr[100],arr2[200][100];

    int h,w; cin>>h>>w;

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

        cin>>arr[i];

    }

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

        cout<<arr[i]<<endl;

        cout<<arr[i]<<endl;

    }

}

// Faster C++ solution for the same problem (measured speedup: 1.662638663731122x):
#include <bits/stdc++.h>

using namespace std;



const int N = 1100;



char pan[N][N];



int main() {

	

	ios::sync_with_stdio(0);

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

	

	int n, m; cin >> n >> m;

	for (int i = 0; i < n; i++) cin >> pan[i];

	for (int i = 0; i < n + n; i++) cout << pan[i / 2] << "\n"; 

	

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

const int N=65;

const double eps=1e-8;

int n,k;

double x[N],y[N],c[N],r[N];

double dis(pair<double,double> a,pair<double,double> b) {

    return sqrt((a.first-b.first)*(a.first-b.first)+(a.second-b.second)*(a.second-b.second));

}

vector<pair<double,double> > cal(double x1,double y1,double r1,double x2,double y2,double r2) {//两圆交点

    x1-=x2,y1-=y2;

    double S=x1*x1+y1*y1,a=(S+r2*r2-r1*r1)/2,D=S*r2*r2-a*a;

    if(S==0)return {};

    if(D<0) return {};

    double A1=a*x1,B1=y1*sqrt(D);

    double A2=a*y1,B2=x1*sqrt(D);

    return {{(A1+B1)/S+x2,(A2-B2)/S+y2},{(A1-B1)/S+x2,(A2+B2)/S+y2}};

}

bool check(double t) {

    for(int i=0;i<n;i++) r[i]=t/c[i];

    vector<pair<double,double> > vec;

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

    	vec.push_back({x[i],y[i]});

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

            auto v=cal(x[i],y[i],r[i],x[j],y[j],r[j]);

            for(auto it:v) vec.push_back(it);

        }

    }

    for(auto it:vec) {

        int cnt=0;

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

            if(r[i]+eps>=dis(it,{x[i],y[i]})) cnt++;

        }

        if(cnt>=k) return true;

    }

    return false;

}

int main() {

    scanf("%d%d",&n,&k);

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

        scanf("%lf%lf%lf",&x[i],&y[i],&c[i]);

    }

    double L=0,R=1000010;

    for(int i=0;i<100;i++) {

        double mid=(L+R)/2;

        if(check(mid)) R=mid;

        else L=mid;

    }

    printf("%.6lf\n",R);

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.492926859225791x):
#include <stdio.h>

#include <math.h>

#define maxn 65

#define eps 1e-8

double x[maxn],y[maxn],c[maxn],X[65*65*2],Y[65*65*2],r[maxn];

int tot,N,K;

double dis(double x,double y,double c,double X,double Y){

	return sqrt((X-x)*(X-x)+(Y-y)*(Y-y));//return c*sqrt((X-x)*(X-x)+(Y-y)*(Y-y));

}

void cal(double x1,double y1,double r1,double x2,double y2,double r2){//计算两圆交点坐标

	double D;//A=a*a,H=h*h,D=d*d;

	double ad,h;//ad=a*d;

	D=(x2-x1)*(x2-x1)+(y2-y1)*(y2-y1);

	if(D==0)return;

	ad=(r1*r1+D-r2*r2)/2;//2*a*d=r1*r1+D-r2*r2

	h=D*r1*r1-ad*ad;

	if(h<0)return;//漏了此句的判定

	h=sqrt(h);

	tot++,X[tot]=x1+ad*(x2-x1)/D-(y2-y1)*h/D,Y[tot]=y1+ad*(y2-y1)/D+(x2-x1)*h/D;

	tot++,X[tot]=x1+ad*(x2-x1)/D+(y2-y1)*h/D,Y[tot]=y1+ad*(y2-y1)/D-(x2-x1)*h/D;

}

int judge(double t){

	int i,j,cnt;

	for(i=1;i<=N;i++)r[i]=t/c[i];

	tot=0;

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

		tot++,X[tot]=x[i],Y[tot]=y[i];//tot++,X[i]=x[i],Y[i]=y[i];

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

			cal(x[i],y[i],r[i],x[j],y[j],r[j]);

	}

	for(i=1;i<=tot;i++){

		cnt=0;

		for(j=1;j<=N;j++)

			if(dis(x[j],y[j],c[j],X[i],Y[i])<r[j]+eps)cnt++;//浮点比大小无等于的情况，这是由浮点误差决定的。if(dis(x[j],y[j],c[j],X[i],Y[i])<t+eps)cnt++;

		if(cnt>=K)return 1;

	}

	return 0;

}

int main(){

	double l,r,mid;

	int i;

	scanf("%d%d",&N,&K);

	for(i=1;i<=N;i++)scanf("%lf%lf%lf",&x[i],&y[i],&c[i]);

	l=0,r=1000010;

	for(i=1;i<=100;i++){

		mid=(l+r)/2;

		if(judge(mid))r=mid;

		else l=mid;

	}

	printf("%.6lf\n",r);

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

const int N=65;

const double eps=1e-8;

int n,k;

double x[N],y[N],c[N],r[N];

double dis(pair<double,double> a,pair<double,double> b) {

    return sqrt((a.first-b.first)*(a.first-b.first)+(a.second-b.second)*(a.second-b.second));

}

vector<pair<double,double> > cal(double x1,double y1,double r1,double x2,double y2,double r2) {//两圆交点

    x1-=x2,y1-=y2;

    double S=x1*x1+y1*y1,a=(S+r2*r2-r1*r1)/2,D=S*r2*r2-a*a;

    if(S==0)return {};

    if(D<0) return {};

    double A1=a*x1,B1=y1*sqrt(D);

    double A2=a*y1,B2=x1*sqrt(D);

    return {{(A1+B1)/S+x2,(A2-B2)/S+y2},{(A1-B1)/S+x2,(A2+B2)/S+y2}};

}

bool check(double t) {

    for(int i=0;i<n;i++) r[i]=t/c[i];

    vector<pair<double,double> > vec;

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

    	vec.push_back({x[i],y[i]});

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

            auto v=cal(x[i],y[i],r[i],x[j],y[j],r[j]);

            for(auto it:v) vec.push_back(it);

        }

    }

    for(auto it:vec) {

        int cnt=0;

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

            if(r[i]+eps>=dis(it,{x[i],y[i]})) cnt++;

        }

        if(cnt>=k) return true;

    }

    return false;

}

int main() {

    scanf("%d%d",&n,&k);

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

        scanf("%lf%lf%lf",&x[i],&y[i],&c[i]);

    }

    double L=0,R=1000010;

    for(int i=0;i<100;i++) {

        double mid=(L+R)/2;

        if(check(mid)) R=mid;

        else L=mid;

    }

    printf("%.6lf\n",R);

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.4758857397471115x):
#include <stdio.h>

#include <math.h>

#define maxn 65

#define eps 1e-8

double x[maxn],y[maxn],c[maxn],X[65*65*2],Y[65*65*2],r[maxn];

int tot,N,K;

double dis(double x,double y,double c,double X,double Y){

	return c*sqrt((X-x)*(X-x)+(Y-y)*(Y-y));

}

void cal(double x1,double y1,double r1,double x2,double y2,double r2){//计算两圆交点坐标

	double D;//A=a*a,H=h*h,D=d*d;

	double ad,h;//ad=a*d;

	D=(x2-x1)*(x2-x1)+(y2-y1)*(y2-y1);

	if(D==0)return;

	ad=(r1*r1+D-r2*r2)/2;//2*a*d=r1*r1+D-r2*r2

	h=D*r1*r1-ad*ad;

	if(h<0)return;//漏了此句的判定

	h=sqrt(h);

	tot++,X[tot]=x1+ad*(x2-x1)/D-(y2-y1)*h/D,Y[tot]=y1+ad*(y2-y1)/D+(x2-x1)*h/D;

	tot++,X[tot]=x1+ad*(x2-x1)/D+(y2-y1)*h/D,Y[tot]=y1+ad*(y2-y1)/D-(x2-x1)*h/D;

}

int judge(double t){

	int i,j,cnt;

	for(i=1;i<=N;i++)r[i]=t/c[i];

	tot=0;

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

		tot++,X[tot]=x[i],Y[tot]=y[i];//tot++,X[i]=x[i],Y[i]=y[i];

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

			cal(x[i],y[i],r[i],x[j],y[j],r[j]);

	}

	for(i=1;i<=tot;i++){

		cnt=0;

		for(j=1;j<=N;j++)

			if(dis(x[j],y[j],c[j],X[i],Y[i])<t+eps)cnt++;//浮点比大小无等于的情况，这是由浮点误差决定的。if(dis(x[j],y[j],c[j],X[i],Y[i])<t+eps)cnt++;

		if(cnt>=K)return 1;

	}

	return 0;

}

int main(){

	double l,r,mid;

	int i;

	scanf("%d%d",&N,&K);

	for(i=1;i<=N;i++)scanf("%lf%lf%lf",&x[i],&y[i],&c[i]);

	l=0,r=1000010;

	for(i=1;i<=100;i++){

		mid=(l+r)/2;

		if(judge(mid))r=mid;

		else l=mid;

	}

	printf("%.6lf\n",r);

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

const int N=65;

const double eps=1e-8;

int n,k;

double x[N],y[N],c[N],r[N];

double dis(pair<double,double> a,pair<double,double> b) {

    return sqrt((a.first-b.first)*(a.first-b.first)+(a.second-b.second)*(a.second-b.second));

}

vector<pair<double,double> > cal(double x1,double y1,double r1,double x2,double y2,double r2) {//两圆交点

    x1-=x2,y1-=y2;

    double S=x1*x1+y1*y1,a=(S+r2*r2-r1*r1)/2,D=S*r2*r2-a*a;

    if(S==0)return {};

    if(D<0) return {};

    double A1=a*x1,B1=y1*sqrt(D);

    double A2=a*y1,B2=x1*sqrt(D);

    return {{(A1+B1)/S+x2,(A2-B2)/S+y2},{(A1-B1)/S+x2,(A2+B2)/S+y2}};

}

bool check(double t) {

    for(int i=0;i<n;i++) r[i]=t/c[i];

    vector<pair<double,double> > vec;

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

    	vec.push_back({x[i],y[i]});

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

            auto v=cal(x[i],y[i],r[i],x[j],y[j],r[j]);

            for(auto it:v) vec.push_back(it);

        }

    }

    for(auto it:vec) {

        int cnt=0;

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

            if(r[i]+eps>=dis(it,{x[i],y[i]})) cnt++;

        }

        if(cnt>=k) return true;

    }

    return false;

}

int main() {

    scanf("%d%d",&n,&k);

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

        scanf("%lf%lf%lf",&x[i],&y[i],&c[i]);

    }

    double L=0,R=1000010;

    for(int i=0;i<100;i++) {

        double mid=(L+R)/2;

        if(check(mid)) R=mid;

        else L=mid;

    }

    printf("%.6lf\n",R);

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.730716862773463x):
#include <stdio.h>

#include <math.h>

#define maxn 65

#define eps 1e-8

double x[maxn],y[maxn],c[maxn],X[65*65*2],Y[65*65*2],r[maxn];

int tot,N,K;

double dis(double x,double y,double c,double X,double Y){

	return c*sqrt((X-x)*(X-x)+(Y-y)*(Y-y));

}

void cal(double x1,double y1,double r1,double x2,double y2,double r2){//计算两圆交点坐标

	double D;//A=a*a,H=h*h,D=d*d;

	double ad,h;//ad=a*d;

	D=(x2-x1)*(x2-x1)+(y2-y1)*(y2-y1);

	if(D==0)return;

	ad=(r1*r1+D-r2*r2)/2;//2*a*d=r1*r1+D-r2*r2

	h=D*r1*r1-ad*ad;

	if(h<0)return;//漏了此句的判定

	h=sqrt(h);

	tot++,X[tot]=x1+ad*(x2-x1)/D-(y2-y1)*h/D,Y[tot]=y1+ad*(y2-y1)/D+(x2-x1)*h/D;

	tot++,X[tot]=x1+ad*(x2-x1)/D+(y2-y1)*h/D,Y[tot]=y1+ad*(y2-y1)/D-(x2-x1)*h/D;

}

int judge(double t){

	int i,j,cnt;

	for(i=1;i<=N;i++)r[i]=t/c[i];

	tot=0;

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

		tot++,X[tot]=x[i],Y[tot]=y[i];//tot++,X[i]=x[i],Y[i]=y[i];

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

			cal(x[i],y[i],r[i],x[j],y[j],r[j]);

	}

	for(i=1;i<=tot;i++){

		cnt=0;

		for(j=1;j<=N;j++)

			if(dis(x[j],y[j],c[j],X[i],Y[i])<t+eps)cnt++;//浮点比大小无等于的情况，这是由浮点误差决定的。

		if(cnt>=K)return 1;

	}

	return 0;

}

int main(){

	double l,r,mid;

	int i;

	scanf("%d%d",&N,&K);

	for(i=1;i<=N;i++)scanf("%lf%lf%lf",&x[i],&y[i],&c[i]);

	l=0,r=1000010;

	while(l+eps<r){

		mid=(l+r)/2;

		if(judge(mid))r=mid;

		else l=mid;

	}

	printf("%.6lf\n",r);

	return 0;

}
// Slower C++ solution:
#include <iostream>

#include <vector>

#include <string>

#include <algorithm>

#include <utility>

using namespace std;

//変数デバッグ

#define DEB(variable) cout << #variable << '=' << variable << endl



//for簡易表記(引数ミス防止)

#define FOR(LoopVariable,numberOFbegin,numberOFend) for (int LoopVariable = (numberOFbegin); (LoopVariable) < (numberOFend); (LoopVariable)++)

#define REP(LoopVariable,numberOFend) for(int LoopVariable = 0;(LoopVariable)<(numberOFend);LoopVariable++)





int main(){



    int n,a,b;

    cin >> n >> a >> b;

    if((b-a)%2){

        cout << "Borys" << endl;

    }else{

        cout << "Alice" << endl;

    }

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.311258915861989x):
#include <stdio.h>



#define endl '\n'



int main()

{

    int n, a, b;

    scanf("%d %d %d", &n, &a, &b);



    a-=b;

    a&1? printf("Borys"): printf("Alice");



    return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;



int main() {

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    cout.tie(NULL);

    

    int N, A, B;

    cin >> N >> A >> B;



    int diff = B - A;



    // CHECK FOR SPECIAL CASE WHEN A IS NEXT TO B

    if(diff == 0 && (A==1 || B==N)) {

        if(A==1) cout << "Borys" << endl;

    }

    else {

        // CHECK PARITY: ALICE WINS IF EVEN

        if(diff%2 == 0) cout << "Alice" << endl;

        else cout << "Borys" << endl;

    }

}

// Faster C++ solution for the same problem (measured speedup: 5.382745857014484x):
#include <stdio.h>



#define endl '\n'



int main()

{

    int n, a, b;

    scanf("%d %d %d", &n, &a, &b);



    a-=b;

    a&1? printf("Borys"): printf("Alice");



    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



int n, a, b;



int main() {

    scanf("%d%d%d", &n, &a, &b);

    if((b - a) % 2 == 0) printf("Alice\n");

    else printf("Borys\n");

}

// Faster C++ solution for the same problem (measured speedup: 5.241147073849919x):
#include <stdio.h>

#include <string.h>

#include <math.h>

int main ()

{

    int n, a, b;

    scanf ("%d %d %d", &n, &a, &b);

    if ((a+b)%2==0) printf ("Alice");

    else printf ("Borys");

    return 0;

}
// Slower C++ solution:
#include <algorithm>

#include <climits>

#include <cmath>

#include <iomanip>

#include <iostream>

#include <list>

#include <map>

#include <set>

#include <sstream>

#include <string>

#include <vector>



#define endl '\n'

#define ull unsigned long long

#define ll long long

#define ul unsigned long

#define ld long double

#define PI 3.14159265359



using namespace std;



int main() {

	ios_base::sync_with_stdio(false);

	cin.tie(nullptr);

	cout.tie(nullptr);

	int t = 1;

	// cin >> t;

	while (t--) {

		int n, a, b;

		cin >> n >> a >> b;

		cout << ((b - a) & 1 ? "Borys" : "Alice");

	}

}

// Faster C++ solution for the same problem (measured speedup: 5.372954452633747x):
#include <stdio.h>

#include <string.h>

#include <math.h>

int main ()

{

    int n, a, b;

    scanf ("%d %d %d", &n, &a, &b);

    if ((a+b)%2==0) printf ("Alice");

    else printf ("Borys");

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



int main()

{

    int n, a, b;

    cin >> n >> a >> b;

    int d = b-a;

    

    if(d%2) cout << "Borys" << endl;

    else cout << "Alice" << endl;



    return 0; 

}

// Faster C++ solution for the same problem (measured speedup: 5.31078223762152x):
#include <stdio.h>

#include <string.h>

#include <math.h>

int main ()

{

    int n, a, b;

    scanf ("%d %d %d", &n, &a, &b);

    if ((a+b)%2==0) printf ("Alice");

    else printf ("Borys");

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

int main()

{

    int N, A, B; cin >> N >> A >> B;

    cout << ((B - A) % 2 ? "Borys" : "Alice") << endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.316579500734389x):
#include<stdio.h>



int a,b;



int main(void){

    scanf("%*d%d%d",&a,&b);

    if((b-a-1)&1)

        puts("Alice");

    else

        puts("Borys");

    return 0;

}
// Slower C++ solution:
#include<iostream>

#include<algorithm>

#include<cstring>

#include<set>

#include<vector>

#include<queue>

#include<cmath>

#include<memory.h>

#include<ctime>

#include<cstdio>

#include<map>

#include<stack>



using namespace std;



int main()

{

	int n,a,b;

	cin>>n>>a>>b;

	if((b-a)%2==0)

	{

		cout<<"Alice"<<endl;

	}

	else

	{

		cout<<"Borys"<<endl;

	}

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.309355735149921x):
#include<stdio.h>



int a,b;



int main(void){

    scanf("%*d%d%d",&a,&b);

    if((b-a-1)&1)

        puts("Alice");

    else

        puts("Borys");

    return 0;

}