I Love You Too

Problem Description

This is a true story. A man showed his love to a girl,but the girl didn't replied clearly ,but gave him a Morse Code:

****-/*----/----*/****-/****-/*----/---**/*----/****-/*----/-****/***--/****-/*----/----*/**---/-****/**---/**---/***--/--***/****-/ He was so anxious that he asked for help in the Internet and after one day a
girl named "Pianyi angel" found the secret of this code. She translate this code as this five steps:

1.First translate the morse code to a number string:4194418141634192622374

2.Second she cut two number as one group 41 94 41 81 41 63 41 92 62 23 74,according to standard Mobile phone can get this alphabet:GZGTGOGXNCS



3.Third she change this alphabet according to the keyboard:QWERTYUIOPASDFGHJKLZXCVBNM = ABCDEFGHIJKLMNOPQRSTUVWXYZ

So ,we can get OTOEOIOUYVL

4.Fourth, divide this alphabet to two parts: OTOEOI and OUYVL, compose again.we will get OOTUOYEVOLI

5.Finally,reverse this alphabet the answer will appear : I LOVE YOU TOO



I guess you might worship Pianyi angel as me,so let's Orz her.

Now,the task is translate the number strings.

Input

A number string each line(length <= 1000). I ensure all input are legal.

Output

An upper alphabet string.

SampleInput

4194418141634192622374
41944181416341926223

SampleOutput

ILOVEYOUTOO
VOYEUOOTIO

／／标程：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
using namespace std;
string s[15], s1, s2;
string a = "QWERTYUIOPASDFGHJKLZXCVBNM";
string b = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
void init()
{
s[2] = "ABC";
s[3] = "DEF";
s[4] = "GHI";
s[5] = "JKL";
s[6] = "MNO";
s[7] = "PQRS";
s[8] = "TUV";
s[9] = "WXYZ";
}
void fun(char ch1,char ch2)
{
int i = ch1 - '0', j = ch2 - '0';
s1 += s[i][j-1];
}
void conversion()
{
for(int i = 0; i < s1.size(); ++ i)
s2 += b[a.find(s1[i])];
}
int main()
{
//   freopen("a.txt","r",stdin);
init();
string str;
int i;
while(cin >> str)
{
s1 = s2 = "";
for(i = 0; i < str.size(); i += 2)
fun(str[i],str[i+1]);
int len = (s1.size() + 1) /2;
conversion();
string s_1 = s2.substr(0,len);
string s_2 = s2.substr(len,s1.size()-len);
string s_3 = "";
for(i = 0; i < s_2.size(); ++ i)
{
s_3 += s_1[i];
s_3 += s_2[i];
}
if(s_1.size() > s_2.size())
s_3 += s_1[i];
reverse(s_3.begin(),s_3.end());
cout << s_3 << endl;
}
return 0;
}

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