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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define LL long long
#define INF 1<<31-1
#define LNF 9223372036854775807
#define PI 3.14159265358979
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define FOR(i,a,n) for(i=a;i<n;i++)
#define MOD 1000000007
#define NUM 100005

void swap(int* a,int* b){int tmp=*a;*a=*b;*b=tmp;}
int abs(int a){if(a<0)return -a;return a;}
int dup(int a,int d){if(a%d==0)return a/d;else return a/d+1;}
int mod(int a,int m){if(a<0)return a%m+m;else return a%m;}
int GCD(int a,int b){int c=1,tmp=MAX(a,b);b=MIN(a,b);a=tmp;while(c!=0){c=a%b;a=b;b=c;}return a;}
int LCM(int a,int b){return a*b/GCD(a,b);}
int POW(int a,int n){LL ans=1;while(n>0){if(n&1)ans=ans*a%MOD;a=a*a%MOD;n>>=1;}return (int)ans;}
int icmp(const void* a,const void* b){return *(int*)a-*(int*)b;}
int lcmp(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return 1;if(*(LL*)a<*(LL*)b)return -1;return 0;}
int ccmp(const void* a,const void* b){return *(char*)a-*(char*)b;}
int scmp(const void* a,const void* b){return strcmp(*(char**)a,*(char**)b);}
LL fac[NUM],finv[NUM],inv[NUM];
void comi(){int i;fac[0]=fac[1]=1;finv[0]=finv[1]=1;inv[1]=1;FOR(i,2,NUM){fac[i]=fac[i-1]*i%MOD;inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;finv[i]=finv[i-1]*inv[i]%MOD;}}
int com(int n,int k){if(n<k||n<0||k<0)return 0;return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;}


int main(){
  int N,i,l,ans=0;
  char S[NUM];
  scanf("%d",&N);
  scanf("%s",S);
  l=strlen(S);
  FOR(i,0,l){
    S[i]+=N;
    if(S[i]>90)S[i]-=26;
  }

  printf("%s\n",S);
	return 0;
}