Problem:
Score : 200 points

Problem Statement

We have a string S consisting of uppercase English letters. Additionally, an integer N will be given.

Shift each character of S by N in alphabetical order (see below), and print the resulting string.

We assume that A follows Z. For example, shifting A by 2 results in C (A \to B \to C), and shifting Y by 3 results in B (Y \to Z \to A \to B).

Constraints

0 \leq N \leq 26

1 \leq |S| \leq 10^4

S consists of uppercase English letters.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the string resulting from shifting each character of S by N in alphabetical order.

Sample Input 1

2
ABCXYZ

Sample Output 1

CDEZAB

Note that A follows Z.

Sample Input 2

0
ABCXYZ

Sample Output 2

ABCXYZ

Sample Input 3

13
ABCDEFGHIJKLMNOPQRSTUVWXYZ

Sample Output 3

NOPQRSTUVWXYZABCDEFGHIJKLM

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/******************************************
* 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;
}