Codeforces Round #424 (Div. 2) B. Keyboard Layouts（字符串处理）

B. Keyboard Layouts

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters
which coincides with English alphabet.

You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.

You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys
were pressed in the second layout.

Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.

Input

The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.

The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.

The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the
text typed in the first layout. The length of s does not exceed 1000.

Output

Print the text if the same keys were pressed in the second layout.

Examples

input

qwertyuiopasdfghjklzxcvbnm
veamhjsgqocnrbfxdtwkylupzi
TwccpQZAvb2017

output

HelloVKCup2017

input

mnbvcxzlkjhgfdsapoiuytrewq
asdfghjklqwertyuiopzxcvbnm
7abaCABAABAcaba7

output

7uduGUDUUDUgudu7

题意：给出两串字符串，一一对应，然后给出一个目标串让你翻译出来。

分析：简单的密码解锁问题，记录一下对应的字符，再记录一下它们的大写，然后对应输出就可以了

AC代码：

#include<stdio.h>
#include<string.h>
#include<map>
using namespace std;
map<char,char>q;
int main()
{
char a[30],b[30];
scanf("%s",a);
scanf("%s",b);
q.clear();
for(int i=0;i<26;i++)
{
q[a[i]]=b[i];
q[a[i]-32]=b[i]-32;
}
char aim[2000];
scanf("%s",aim);
for(int i=0;aim[i];i++)
{
if(q[aim[i]])
printf("%c",q[aim[i]]);
else
printf("%c",aim[i]);
}
printf("\n");
}