#include<stdio.h>
#define N 85
int k1, k2;
char plain[N], cipher[N];
void init(char s[]) //预处理，转换为大写字母
{int i;for(i = 0; s[i]; i++)if(s[i] >= 'a' && s[i] <= 'z')s[i] = s[i] - 'a' + 'A';
}
void encrypt(char s[],int k,int f)//加解密
{init(s);int i;for(i = 0; s[i]; i++)if(s[i] >= 'A' && s[i] <= 'Z')printf("%c", ((s[i] - 'A' + k*f + 26) % 26  + 'A'));printf("\n");
}
int main()
{scanf("%d,%d ", &k1, &k2);gets(plain);gets(cipher);encrypt(plain,k1,1);encrypt(cipher,k2,-1);return 0;
}

#include<stdio.h>
#include<string.h>
#define N 1000
#define ll long long
ll n, e;
char plain[N];
ll qpow(ll a, ll b)//快速幂
{ll k = a % n;ll ans = 1;while(b){if(b & 1)ans = (ans * k ) % n;k = ( k * k) % n;b >>= 1;}return ans;
}
void init(char s[]) //预处理，转换为大写字母
{int i;for(i = 0; s[i]; i++)if(s[i] >= 'a' && s[i] <= 'z')s[i] = s[i] - 'a' + 'A';
}
void encrypt(char s[]) //加密
{init(s);int k = 0,i;ll block = 0;for(i = 0; s[i]; i++)if(s[i] >= 'A' && s[i] <= 'Z') //如果是字母
        {block = block * 100 + (s[i] - 'A') ; //明文对应的数字串k++;if(k == 4) //够8位数字时
            {printf("%08lld", qpow(block, e) % 100000000);//输出密文，这个模不知道是不是必须的，题目没说n的上限block = 0;k = 0;}}if(k)//剩下的明文要后面补零
    {while(k != 4){block = block * 100;k++;}printf("%08lld", qpow(block, e) % 100000000);}
}
int main()
{//  freopen("in.txt", "r", stdin);scanf("%lld%lld ", &n, &e);gets(plain);encrypt(plain);return 0;
}

#include<stdio.h>
#include<string.h>
#define N 802
#define ll long long
ll n, e, p, q, d, x, y;
char cipher[N << 1], plain[10];
ll exgcd(ll a, ll b)//扩展欧几里德求逆元
{if(b == 0){x = 1;y = 0;return a;}ll r = exgcd(b, a % b);ll tmp = x;x = y;y = tmp - a / b * y;return r;
}
ll qpow(ll a, ll b)//快速幂
{ll k = a % n;ll ans = 1;while(b){if(b & 1)ans = (ans * k ) % n;k = ( k * k) % n;b >>= 1;}return ans;
}
void init()//求p和q，和d
{int i;for(i = 2; i < n; i++)if(n % i == 0){p = i;q = n / i;break;}ll M;M = (p - 1) * (q - 1);exgcd(e, M);d = (x % M + M) % M;//求e的逆元d
}
void decrypt() //解密
{ll block = 0;int k = 0, i, j;for(i = 0; cipher[i]; i++){block = block * 10 + cipher[i] - '0';//密文从字符串中取出来k++;if(k == 8)//密文达到8位数字时
        {memset(plain, 0, sizeof plain);//清空明文字符串block = qpow(block, d);//计算明文数字串for(j = 3; j >= 0; j--)//每次计算两位数字并存在明文字符串中，因为从后面往前取，所以倒过来存
            {plain[j] = block % 100 + 'A';//取最后面两个数字block /= 100;//去掉最后面两个数字
            }printf("%s", plain);//输出明文k = 0;//清空计数器
        }}
}
int main()
{//  freopen("in.txt", "r", stdin);scanf("%lld%lld ", &n, &e);gets(cipher);init();printf("%d\n", d);decrypt();return 0;
}