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++ (GCC 9.2.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
#include "ext/pb_ds/tree_policy.hpp"
#include "ext/pb_ds/assoc_container.hpp"
using namespace std;
////////////// Prewritten code follows. Look down for solution. ////////////////
#define x first
#define y second
#define LEN(x) ((int)(x).size())
#define ALL(x) x.begin(), x.end()
using ll = long long;
using llu = long long unsigned;
using ld = long double;
using pii = pair<int, int>;
using vi = vector<int>;
using vpii = vector<pii>;

const ld EPS = 1e-9;
inline int cmp(ld x, ld y = 0, ld tol = EPS) {
	return (x <= y + tol) ? (x + tol < y) ? -1 : 0 : 1;
}

const int MOD = 1;
inline int mod(ll x, int m = MOD) {
	return (int)(((x%m) + m)%m);
}

template<typename T, typename M = __gnu_pbds::null_type>
using ordered_set = __gnu_pbds::tree<T, M, less<T>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update>;

////////////////////////// Solution starts below. //////////////////////////////



int main () {
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.precision(10);
	int n;
	string s;
	cin >> n >> s;
	for(int i = 0; i < LEN(s); i++) {
		int val = s[i]-'A';
		val = (val+n)%26;
		s[i] = val+'A';
	}
	cout << s << '\n';

	return 0;
}