stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
/* ANTHOR KUMAR DAS

KUET ECE 2K18*/

#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define sc(a) scanf("%d",&a)

#define sc1(n) scanf("%lld",&n)

#define sc2(b,c) scanf("%lld%lld",&b,&c)

#define f(a,b) for(a=0;a<b;a++)

#define pb(a) push_back(a)

#define br break

int main()

{

    int n,i,k;

    char j;

    cin>>n;

    char s[90000];

    scanf("%s",s);

    k=strlen(s);

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

    {

        if(s[i]+n<=90)

        {

            j=s[i]+n;

            cout<<j;

        }

        else{

            j=s[i]+n-26;

            cout<<j;

        }

    }



    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.255614739628009x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	char inp[10001];

	scanf("%s", inp);



	for(int i=0; inp[i]!='\0';i++){

		if(inp[i]+n>'Z'){

			inp[i]=inp[i]+n-26;

		}else{

			inp[i]+=n;

		}

	}

	printf("%s", inp);

	return 0;

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

using namespace std;

typedef long long int ll;

///                    BISMILLAHIR RAHMANIR RAHIM



int main()

{

   int n;

   cin>>n;

   string s;

   cin>>s;

   int l=s.length();

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

    int r=s[i]+n;

    if(r>90)

        r=r-26;

    printf("%c",(r));

   }

   cout<<endl;



}







/// A L H A M D U L I L L A H


// Faster C++ solution for the same problem (measured speedup: 5.290621716547495x):
#include <stdio.h>



int main(){

	int n;

	scanf("%d", &n);

	char inp[10001];

	scanf("%s", inp);



	for(int i=0; inp[i]!='\0';i++){

		if(inp[i]+n>'Z'){

			inp[i]=inp[i]+n-26;

		}else{

			inp[i]+=n;

		}

	}

	printf("%s", inp);

	return 0;

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

#include <cstring>

using namespace std;

int main()

{

	int N;

	char a[10005];

	cin>>N>>a;

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

	{

		a[i]+=N;

		if(a[i]>'Z')

			a[i]-=26;

	}

	cout<<a;

} 

// Faster C++ solution for the same problem (measured speedup: 5.202453143610706x):
#include <stdio.h>

#include <string.h>



int main (){

	char huruf[10000];

	int input;

	scanf("%d",&input);

	getchar();

	scanf ("%[^\n]",&huruf);

	int len = strlen (huruf);

	input = input % 26;

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

		if (huruf[i]-'A' + input >= 26){

			huruf[i] = ((huruf[i]-'A' + input) % 26 )+'A';

		} else {

			huruf [i] +=input; 

		}

		printf("%c", huruf[i]);

	}

	

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



using namespace std;



int ans[10000 + 5];



int main(){

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

	

	int n;

	cin >> n;

	string s;

	cin >> s;

	for(int i = 1; i <= (int)s.size(); i++){

		int x = (int) s[i - 1];

		x -= 65;

		x += n;

		x %= 26;

		x += 65;

		ans[i] = x;

	}

	for(int i = 1; i <= (int)s.size(); i++)

		cout << (char) ans[i];

	

	return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.332031077581278x):
#include <stdio.h>

#include <string.h>



int main (){

	char huruf[10000];

	int input;

	scanf("%d",&input);

	getchar();

	scanf ("%[^\n]",&huruf);

	int len = strlen (huruf);

	input = input % 26;

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

		if (huruf[i]-'A' + input >= 26){

			huruf[i] = ((huruf[i]-'A' + input) % 26 )+'A';

		} else {

			huruf [i] +=input; 

		}

		printf("%c", huruf[i]);

	}

	

}
// Slower C++ solution:
/******************************************

* AUTHOR : RAJAGOPALAN *

* NICK : ARNO *

* INSTITUTION : VIT *

******************************************/

#include <bits/stdc++.h>

#include <ext/pb_ds/detail/standard_policies.hpp>

#define ll long long

#define ull unsigned long long

#define pb push_back

#define mp make_pair

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

#define alli(a, n, k) (a+k),(a+n+k)

#define FP(i, a, b, k) for(__typeof(a) i = a;i < b;i += k)

#define FS(i, a, b, k) for(__typeof(a) i = a;i > b;i -= k)

#define IT(it, a) for(__typeof(a.begin()) it = a.begin();it != a.end(); ++it)

#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>

#define eps 1e-6

#define pi 3.141592653589793

using namespace std;

using namespace __gnu_pbds;



template<class T> inline T gcd(T x, T y) { if (!y) return x; return gcd(y, x%y);}

typedef vector<int> VII;

typedef vector<ll> VLL;

typedef pair<int, int> PII;

typedef vector< pair<int, int> > VPII;

typedef vector< pair<int, PII > > VPPI;

const int MOD = 1e9 + 7;

const int INF = 1e9;



inline ll modulo(ll a, ll m)

{

    return (a % m + m) % m;

}



inline ll modInverse(ll a, ll m)

{

    assert(__gcd(a, m) == 1);

    ll m0 = m;

    ll y = 0, x = 1;



    if (m == 1)

        return 0;



    while (a > 1)

    {

        ll q = a / m;

        ll t = m;



        m = a % m, a = t;

        t = y;



        y = x - q * y;

        x = t;

    }



    if (x < 0)

        x += m0;



    return x;

}



inline ll modPow(ll x, ll y, ll m)

{ //x^y % m

    if (y == 0)

        return 1LL;

    else if (y == 1)

        return x;

    else

    {

        ll ans = modPow(x, y / 2, m) % m;

        if (y & 1)

        {

            return (((ans * ans) % m) * x) % m;

        }

        else

        {

            return (ans * ans) % m;

        }

    }

}



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

{

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    int n;

    string s;

    cin>>n>>s;

    for(int i=0;i<s.size();++i)

    {

        s[i]=(s[i]-'A'+n)%26 +'A';

    }

    cout<<s;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.283929578862184x):
#include <stdio.h>

#include <string.h>



int main (){

	char huruf[10000];

	int input;

	scanf("%d",&input);

	getchar();

	scanf ("%[^\n]",&huruf);

	int len = strlen (huruf);

	input = input % 26;

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

		if (huruf[i]-'A' + input >= 26){

			huruf[i] = ((huruf[i]-'A' + input) % 26 )+'A';

		} else {

			huruf [i] +=input; 

		}

		printf("%c", huruf[i]);

	}

	

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

using namespace std;

int main()

{

	int n,i,cap;

	string s;

	cin>>n;

	cin>>s;

	for(i=0;i<s.size();i++){

		cap=s[i]+n;

		if(cap<=90){

			s[i]=cap;

		}

		else if(cap>90){

			s[i]=cap-26;

		}

	}

	cout<<s<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.27537491813638x):
#include <stdio.h>

#include <string.h>



int main (){

	char huruf[10000];

	int input;

	scanf("%d",&input);

	getchar();

	scanf ("%[^\n]",&huruf);

	int len = strlen (huruf);

	input = input % 26;

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

		if (huruf[i]-'A' + input >= 26){

			huruf[i] = ((huruf[i]-'A' + input) % 26 )+'A';

		} else {

			huruf [i] +=input; 

		}

		printf("%c", huruf[i]);

	}

	

}
// Slower C++ solution:
/* ANTHOR KUMAR DAS

KUET ECE 2K18*/

#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define sc(a) scanf("%d",&a)

#define sc1(n) scanf("%lld",&n)

#define sc2(b,c) scanf("%lld%lld",&b,&c)

#define f(a,b) for(a=0;a<b;a++)

#define pb(a) push_back(a)

#define br break

int main()

{

    int n,i,k;

    char j;

    cin>>n;

    char s[90000];

    scanf("%s",s);

    k=strlen(s);

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

    {

        if(s[i]+n<=90)

        {

            j=s[i]+n;

            cout<<j;

        }

        else{

            j=s[i]+n-26;

            cout<<j;

        }

    }



    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.203778651425951x):
#include <stdio.h>

#include <string.h>



int main (){

	char huruf[10000];

	int input;

	scanf("%d",&input);

	getchar();

	scanf ("%[^\n]",&huruf);

	int len = strlen (huruf);

	input = input % 26;

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

		if (huruf[i]-'A' + input >= 26){

			huruf[i] = ((huruf[i]-'A' + input) % 26 )+'A';

		} else {

			huruf [i] +=input; 

		}

		printf("%c", huruf[i]);

	}

	

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

using namespace std;

typedef long long int ll;

///                    BISMILLAHIR RAHMANIR RAHIM



int main()

{

   int n;

   cin>>n;

   string s;

   cin>>s;

   int l=s.length();

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

    int r=s[i]+n;

    if(r>90)

        r=r-26;

    printf("%c",(r));

   }

   cout<<endl;



}







/// A L H A M D U L I L L A H


// Faster C++ solution for the same problem (measured speedup: 5.238440354798348x):
#include <stdio.h>

#include <string.h>



int main (){

	char huruf[10000];

	int input;

	scanf("%d",&input);

	getchar();

	scanf ("%[^\n]",&huruf);

	int len = strlen (huruf);

	input = input % 26;

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

		if (huruf[i]-'A' + input >= 26){

			huruf[i] = ((huruf[i]-'A' + input) % 26 )+'A';

		} else {

			huruf [i] +=input; 

		}

		printf("%c", huruf[i]);

	}

	

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

#include <cstring>

using namespace std;

int main()

{

	int N;

	char a[10005];

	cin>>N>>a;

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

	{

		a[i]+=N;

		if(a[i]>'Z')

			a[i]-=26;

	}

	cout<<a;

} 

// Faster C++ solution for the same problem (measured speedup: 5.217799670035486x):




#include<stdio.h>

#include<string.h>

int main()

{

    int n,j,c;

    scanf("%d",&n);

    char s[10000];

    scanf("%s",&s);

    j=strlen(s);

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

        c = s[i] - 'A';

        s[i] = 'A' + (c+n)%26;



}

printf("%s",s);

return 0;

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



using namespace std;



int ans[10000 + 5];



int main(){

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

	

	int n;

	cin >> n;

	string s;

	cin >> s;

	for(int i = 1; i <= (int)s.size(); i++){

		int x = (int) s[i - 1];

		x -= 65;

		x += n;

		x %= 26;

		x += 65;

		ans[i] = x;

	}

	for(int i = 1; i <= (int)s.size(); i++)

		cout << (char) ans[i];

	

	return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.3477598412185525x):




#include<stdio.h>

#include<string.h>

int main()

{

    int n,j,c;

    scanf("%d",&n);

    char s[10000];

    scanf("%s",&s);

    j=strlen(s);

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

        c = s[i] - 'A';

        s[i] = 'A' + (c+n)%26;



}

printf("%s",s);

return 0;

}
// Slower C++ solution:
/******************************************

* AUTHOR : RAJAGOPALAN *

* NICK : ARNO *

* INSTITUTION : VIT *

******************************************/

#include <bits/stdc++.h>

#include <ext/pb_ds/detail/standard_policies.hpp>

#define ll long long

#define ull unsigned long long

#define pb push_back

#define mp make_pair

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

#define alli(a, n, k) (a+k),(a+n+k)

#define FP(i, a, b, k) for(__typeof(a) i = a;i < b;i += k)

#define FS(i, a, b, k) for(__typeof(a) i = a;i > b;i -= k)

#define IT(it, a) for(__typeof(a.begin()) it = a.begin();it != a.end(); ++it)

#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>

#define eps 1e-6

#define pi 3.141592653589793

using namespace std;

using namespace __gnu_pbds;



template<class T> inline T gcd(T x, T y) { if (!y) return x; return gcd(y, x%y);}

typedef vector<int> VII;

typedef vector<ll> VLL;

typedef pair<int, int> PII;

typedef vector< pair<int, int> > VPII;

typedef vector< pair<int, PII > > VPPI;

const int MOD = 1e9 + 7;

const int INF = 1e9;



inline ll modulo(ll a, ll m)

{

    return (a % m + m) % m;

}



inline ll modInverse(ll a, ll m)

{

    assert(__gcd(a, m) == 1);

    ll m0 = m;

    ll y = 0, x = 1;



    if (m == 1)

        return 0;



    while (a > 1)

    {

        ll q = a / m;

        ll t = m;



        m = a % m, a = t;

        t = y;



        y = x - q * y;

        x = t;

    }



    if (x < 0)

        x += m0;



    return x;

}



inline ll modPow(ll x, ll y, ll m)

{ //x^y % m

    if (y == 0)

        return 1LL;

    else if (y == 1)

        return x;

    else

    {

        ll ans = modPow(x, y / 2, m) % m;

        if (y & 1)

        {

            return (((ans * ans) % m) * x) % m;

        }

        else

        {

            return (ans * ans) % m;

        }

    }

}



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

{

    ios::sync_with_stdio(false);

    cin.tie(NULL);

    int n;

    string s;

    cin>>n>>s;

    for(int i=0;i<s.size();++i)

    {

        s[i]=(s[i]-'A'+n)%26 +'A';

    }

    cout<<s;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.29951644964607x):




#include<stdio.h>

#include<string.h>

int main()

{

    int n,j,c;

    scanf("%d",&n);

    char s[10000];

    scanf("%s",&s);

    j=strlen(s);

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

        c = s[i] - 'A';

        s[i] = 'A' + (c+n)%26;



}

printf("%s",s);

return 0;

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

using namespace std;

int main()

{

	int n,i,cap;

	string s;

	cin>>n;

	cin>>s;

	for(i=0;i<s.size();i++){

		cap=s[i]+n;

		if(cap<=90){

			s[i]=cap;

		}

		else if(cap>90){

			s[i]=cap-26;

		}

	}

	cout<<s<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.290936553839188x):




#include<stdio.h>

#include<string.h>

int main()

{

    int n,j,c;

    scanf("%d",&n);

    char s[10000];

    scanf("%s",&s);

    j=strlen(s);

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

        c = s[i] - 'A';

        s[i] = 'A' + (c+n)%26;



}

printf("%s",s);

return 0;

}
// Slower C++ solution:
/* ANTHOR KUMAR DAS

KUET ECE 2K18*/

#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define sc(a) scanf("%d",&a)

#define sc1(n) scanf("%lld",&n)

#define sc2(b,c) scanf("%lld%lld",&b,&c)

#define f(a,b) for(a=0;a<b;a++)

#define pb(a) push_back(a)

#define br break

int main()

{

    int n,i,k;

    char j;

    cin>>n;

    char s[90000];

    scanf("%s",s);

    k=strlen(s);

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

    {

        if(s[i]+n<=90)

        {

            j=s[i]+n;

            cout<<j;

        }

        else{

            j=s[i]+n-26;

            cout<<j;

        }

    }



    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.2191290879178x):




#include<stdio.h>

#include<string.h>

int main()

{

    int n,j,c;

    scanf("%d",&n);

    char s[10000];

    scanf("%s",&s);

    j=strlen(s);

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

        c = s[i] - 'A';

        s[i] = 'A' + (c+n)%26;



}

printf("%s",s);

return 0;

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

using namespace std;

typedef long long int ll;

///                    BISMILLAHIR RAHMANIR RAHIM



int main()

{

   int n;

   cin>>n;

   string s;

   cin>>s;

   int l=s.length();

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

    int r=s[i]+n;

    if(r>90)

        r=r-26;

    printf("%c",(r));

   }

   cout<<endl;



}







/// A L H A M D U L I L L A H


// Faster C++ solution for the same problem (measured speedup: 5.253893038582397x):




#include<stdio.h>

#include<string.h>

int main()

{

    int n,j,c;

    scanf("%d",&n);

    char s[10000];

    scanf("%s",&s);

    j=strlen(s);

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

        c = s[i] - 'A';

        s[i] = 'A' + (c+n)%26;



}

printf("%s",s);

return 0;

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

using namespace std;



typedef long long ll;



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

#define file() freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);

#define tt int t; cin>>t; while(t--)

#define nl cout<<"\n";

#define sp cout<<" ";

#define rep(i, a, b) for(long long i=a; i<b; i++)



const int INF = 1e9 + 7;

const int MOD = 1e9 + 7;

double PI = 4*atan(1);



int main()

{

    optimize();



    int n;

    string s;

    cin>> n>> s;



    for( int i=0; i<s.size(); i++)

    {

        s[i]=( s[i] + n <= 'Z' ? s[i]+ n : s[i]+ n-26);

    }



    cout << s << '\n';



    return 0;

}






// Faster C++ solution for the same problem (measured speedup: 5.298731610041565x):




#include<stdio.h>

#include<string.h>

int main()

{

    int n,j,c;

    scanf("%d",&n);

    char s[10000];

    scanf("%s",&s);

    j=strlen(s);

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

        c = s[i] - 'A';

        s[i] = 'A' + (c+n)%26;



}

printf("%s",s);

return 0;

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

#include <bits/stdc++.h>

using namespace std;

int n,a;

string s;

char c;

int main()

{

    cin >> n >> s;

    for(int i=0;i<s.length();i++){

        if(static_cast<char>(s[i] + n)<'A'||static_cast<char>(s[i] + n)>'Z') cout << static_cast<char>(s[i] -26+ n);

       else cout << static_cast<char>(s[i] + n);

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.045764404209365x):
#include <stdio.h>

#include <string.h>

int main ()

{

	int N,i,c;

	scanf("%d",&N);

	

	char s[100000];

	scanf("%s",s);

	

	

	for(i=0;i<strlen(s);i++)

	{

		c=s[i]+N;

		if(c<=90)

		{

			s[i]=c;

		}

		if(c>90)

		{

			s[i]=c-26;

		}

	}

	printf("%s\n",s);

	return 0;

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



using namespace std;



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

{

    int n;

    string s, s1 = "";

    cin >> n >> s;

    for (char i : s)

    {

        int p1, p2, t;

        t = n;

        p1 = (int)i;

        p2 = min(90, p1 + t);

        t -= (p2 - p1);

        if (t > 0)

            p2 = 65 + t - 1;

        s1.push_back((char(p2)));

    }

    cout << s1;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.039183238118243x):
#include <stdio.h>

#include <string.h>

int main ()

{

	int N,i,c;

	scanf("%d",&N);

	

	char s[100000];

	scanf("%s",s);

	

	

	for(i=0;i<strlen(s);i++)

	{

		c=s[i]+N;

		if(c<=90)

		{

			s[i]=c;

		}

		if(c>90)

		{

			s[i]=c-26;

		}

	}

	printf("%s\n",s);

	return 0;

}
// Slower C++ solution:


#include<vector>

#include<iostream>



using namespace std;



int main(){

    int n; 

    cin>>n;

    string s;

    cin>>s;

    for( int i =0; i < s.size(); i++)

    {

        int temp = s[i] - 65;

        temp = (temp + n ) %26;

        temp += 65;

        s[i] = temp;

    }

    cout<<s;





  



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.077464944964147x):
#include <stdio.h>

#include <string.h>

int main ()

{

	int N,i,c;

	scanf("%d",&N);

	

	char s[100000];

	scanf("%s",s);

	

	

	for(i=0;i<strlen(s);i++)

	{

		c=s[i]+N;

		if(c<=90)

		{

			s[i]=c;

		}

		if(c>90)

		{

			s[i]=c-26;

		}

	}

	printf("%s\n",s);

	return 0;

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

#include <bits/stdc++.h>

using namespace std;

int n,a;

string s;

char c;

int main()

{

    cin >> n >> s;

    for(int i=0;i<s.length();i++){

        if(static_cast<char>(s[i] + n)<'A'||static_cast<char>(s[i] + n)>'Z') cout << static_cast<char>(s[i] -26+ n);

       else cout << static_cast<char>(s[i] + n);

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.3061737838062975x):
#include <stdio.h>

#include <string.h>



int main() {

	int n;

	char s[10005];



	scanf("%d", &n);

	scanf("%s", s);



	for(int i = 0; s[i] != '\0'; i++) {

		s[i] = 'A' + (s[i] - 'A' + n) % 26;

	}



	puts(s);



	return 0;

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



using namespace std;



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

{

    int n;

    string s, s1 = "";

    cin >> n >> s;

    for (char i : s)

    {

        int p1, p2, t;

        t = n;

        p1 = (int)i;

        p2 = min(90, p1 + t);

        t -= (p2 - p1);

        if (t > 0)

            p2 = 65 + t - 1;

        s1.push_back((char(p2)));

    }

    cout << s1;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.299252967021738x):
#include <stdio.h>

#include <string.h>



int main() {

	int n;

	char s[10005];



	scanf("%d", &n);

	scanf("%s", s);



	for(int i = 0; s[i] != '\0'; i++) {

		s[i] = 'A' + (s[i] - 'A' + n) % 26;

	}



	puts(s);



	return 0;

}
// Slower C++ solution:


#include<vector>

#include<iostream>



using namespace std;



int main(){

    int n; 

    cin>>n;

    string s;

    cin>>s;

    for( int i =0; i < s.size(); i++)

    {

        int temp = s[i] - 65;

        temp = (temp + n ) %26;

        temp += 65;

        s[i] = temp;

    }

    cout<<s;





  



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.339510373629077x):
#include <stdio.h>

#include <string.h>



int main() {

	int n;

	char s[10005];



	scanf("%d", &n);

	scanf("%s", s);



	for(int i = 0; s[i] != '\0'; i++) {

		s[i] = 'A' + (s[i] - 'A' + n) % 26;

	}



	puts(s);



	return 0;

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



using namespace std;



int main(){

    string s,c;

    int n,i,j,k;

    cin>>n>>s;

    int len=s.size();

    for(i=0;i<len;i++){

        if(s[i]+n>'Z'){cout<<char(s[i]+n-26);}

        else {cout<<char(s[i]+n);}

    }



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.345228598077018x):
#include <stdio.h>

#include <string.h>



int main() {

	int n;

	char s[10005];



	scanf("%d", &n);

	scanf("%s", s);



	for(int i = 0; s[i] != '\0'; i++) {

		s[i] = 'A' + (s[i] - 'A' + n) % 26;

	}



	puts(s);



	return 0;

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

using namespace std;

int main(){



    int n;

    cin >> n;



    string s;

    cin >> s;



    for(int i=0;i<s.length();i++){

        s[i] = (s[i] - 'A' + n) % 26 + 'A';

    }



    cout << s << endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.362298787965981x):
#include <stdio.h>

#include <string.h>



int main() {

	int n;

	char s[10005];



	scanf("%d", &n);

	scanf("%s", s);



	for(int i = 0; s[i] != '\0'; i++) {

		s[i] = 'A' + (s[i] - 'A' + n) % 26;

	}



	puts(s);



	return 0;

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

#include <bits/stdc++.h>

using namespace std;

int n,a;

string s;

char c;

int main()

{

    cin >> n >> s;

    for(int i=0;i<s.length();i++){

        if(static_cast<char>(s[i] + n)<'A'||static_cast<char>(s[i] + n)>'Z') cout << static_cast<char>(s[i] -26+ n);

       else cout << static_cast<char>(s[i] + n);

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.2650178157509755x):
#include <stdio.h>

#include <string.h>



int main (){



    int n;

    scanf("%d", &n);

    char str[10000];

    scanf("%s", str);

    for(int i = 0; i < strlen(str); i++){

        str[i] = str[i] + n;

        if(str[i] >= 91) str[i] -= 26;

    }



    printf("%s", str);





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



using namespace std;



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

{

    int n;

    string s, s1 = "";

    cin >> n >> s;

    for (char i : s)

    {

        int p1, p2, t;

        t = n;

        p1 = (int)i;

        p2 = min(90, p1 + t);

        t -= (p2 - p1);

        if (t > 0)

            p2 = 65 + t - 1;

        s1.push_back((char(p2)));

    }

    cout << s1;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.258150678496357x):
#include <stdio.h>

#include <string.h>



int main (){



    int n;

    scanf("%d", &n);

    char str[10000];

    scanf("%s", str);

    for(int i = 0; i < strlen(str); i++){

        str[i] = str[i] + n;

        if(str[i] >= 91) str[i] -= 26;

    }



    printf("%s", str);





}
// Slower C++ solution:


#include<vector>

#include<iostream>



using namespace std;



int main(){

    int n; 

    cin>>n;

    string s;

    cin>>s;

    for( int i =0; i < s.size(); i++)

    {

        int temp = s[i] - 65;

        temp = (temp + n ) %26;

        temp += 65;

        s[i] = temp;

    }

    cout<<s;





  



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.298095838915044x):
#include <stdio.h>

#include <string.h>



int main (){



    int n;

    scanf("%d", &n);

    char str[10000];

    scanf("%s", str);

    for(int i = 0; i < strlen(str); i++){

        str[i] = str[i] + n;

        if(str[i] >= 91) str[i] -= 26;

    }



    printf("%s", str);





}