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.*;
import java.util.Map.Entry;
 

public class Main{
	
	
	
	
	public static long answer = 0;
	public static long MOD=1000000007;
	
	
	

	

 
	
	//最大公約数・最小公倍数（セットで使う）
	public static long gcm(long a, long b) {
		long result = a;
		long k = 0;
		long n = b;
		do {
			k = result % n;
			result = n;
			n = k;
		} while(k != 0);
		return result;
	}
	
	public static long lcm(long a, long b) {
		long g;
		g = gcm(a, b);
		return a/g*b;
	}
	
	static long gcd(long a, long b) {
		if (b == 0) {
			return a;			
		}
		return gcd(b, a % b);
	}
	
	//等差数列
	static long tousasum(int n) {
		return (long)n*((long)n+1)/2;
	}
 
 
 
	//素数判定
	static boolean isPrime (int n) {
		if (n==2) return true;
		if (n<2 || n%2==0) return false;
		double d = Math.sqrt(n);
		for (int i=3; i<=d; i+=2) if(n%i==0){return false;}
		return true;
	}
 
 
	//倍数判定（10進数以外のときに有用）
	static boolean isMultiple (String s, int base, int m) {
		int temp = 0;
		for (int i=0; i<s.length(); i++) {
			temp = (temp*base+Character.getNumericValue(s.charAt(i)))%m;
		}
		if (temp==0) {return true;}
		return false;
	}
 
 
	//階乗
	static long factorial (int k) {
		long sum = 1;
		for(int i=1;i<=k;i++) {
			sum = (sum*i)%MOD;
		}
		return sum;
		
	}
 
 
	//進数変換
	static String toNbase (String sm, int m, int n) {
		return Long.toString(Long.parseLong(sm,m),n);
	}
	
	public static final long gcd_n(long[] param) {
	    final int len = param.length;
	    long g = gcd(param[0], param[1]);    //gcd(a, b) は以前作ったもの
	    for (int i = 1; i < len - 1; i++) {
	        g = gcd(g, param[i + 1]);       //gcd(a, b) は以前作ったもの
	    }
	    return g;
	}
	
	public static long lcm_n(long[] numbers) {
		long l;
		l = numbers[0];
		for (int i = 1; i < numbers.length; i++) {
			l = lcm(l, numbers[i]);
		}
		return l;
	}
	
	
    
	
	
 public static void main(String[] args) {
	 
	 Scanner sc=new Scanner(System.in);
	 int N=sc.nextInt();
	 char[] c=sc.next().toCharArray();
	
	 for(int i=0;i<c.length;i++) {
		 int a=c[i];
		 a += N;
		 if(a > 90) {
			 a = 64 + (a - 90);
		 }
		 char b = (char)a;
		 System.out.print(b);
	 }
	 
	 
	 

	 
	 
	
}
 
 
}