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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;
public class Main{
  public static void main(String[] args){
    // 入力
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    String s = sc.next();

    // 準備
    // A~Zを配列(c_ans)に設定
    char[] c_ans = new char[26];
    for(int i=0; i<26; ++i)
      c_ans[i] = (char)('A'+i);

    // 解く
    // c_ans[]に A~Zを0~25 に変換したものを設定し出力する方針
    char[] c = new char[s.length()];
    for(int i=0; i<s.length(); ++i) {

      // 入力sを一文字ずつ「0~25」の値に変換
      c[i] = s.charAt(i);
      int c_num = c[i] - 'A' + n;
      if(c_num > 25) c_num = c_num-26; // 26以上になったらマイナス26。26番目=0番目, 27番目=1番目...

      // 出力
      System.out.print(c_ans[c_num]);
    }
  }
}
// b問題だから弱いと思ったけど、強かった。