As an Android developer, itˇs really not easy to figure out a newer version of two kernels, because Android is updated so frequently and has many branches. Fortunately, Google identifies individual builds with a short build code, e.g. FRF85B.

The first letter is the code name of the release family, e.g. F is Froyo. The code names are ordered alphabetically. The latest code name is K (KitKat).
The second letter is a branch code that allows Google to identify the exact code branch that the build was made from, and R is by convention the primary release branch.

The next letter and two digits are a date code. The letter counts quarters, with A being Q1 2009. Therefore, F is Q2 2010. The two digits count days within the quarter, so F85 is June 24 2010.

Finally, the last letter identifies individual versions related to the same date code, sequentially starting with A; A is actually implicit and usually omitted for brevity.

Please develop a program to compare two Android build numbers.

The first line is an integer n (1 <= n <= 2000), which indicates how many test cases need to process.

Each test case consists of a single line containing two build numbers, separated by a space character.

For each test case, output a single line starting with ¨Case #: 〃 (# means the number of the test case). Then, output the result of release comparison as follows:

● Print "<" if the release of the first build number is lower than the second one;
● Print "=" if the release of the first build number is same as he second one;
● Print ">" if the release of the first build number is higher than the second one.

Continue to output the result of date comparison as follows:
● Print "<" if the date of the first build number is lower than the second one;
● Print "=" if the date of the first build number is same as he second one;
● Print ">" if the date of the first build number is higher than the second one.

If two builds are not in the same code branch, just compare the date code; if they are in the same code branch, compare the date code together with the individual version.

2
FRF85B EPF21B
KTU84L KTU84M

Case 1: > >
Case 2: = <

2014上海全国邀请赛——题目重现（感谢上海大学提供题目）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

char s1[10],s2[10];
char ONE,TWO;

void BJ1()
{
    if(s1[0]<s2[0]) ONE='<';
    else if(s1[0]==s2[0]) ONE='=';
    else ONE='>';
}

void BJ2()
{
    if(s1[2]<s2[2]) TWO='<';
    else if(s1[2]==s2[2])
    {
        int day1=(s1[3]-'0')*10+s1[4]-'0';
        int day2=(s2[3]-'0')*10+s2[4]-'0';
        if(day1<day2) TWO='<';
        else if(day1==day2)
        {
            if(s1[1]==s2[1])
            {
                if(s1[5]<s2[5]) TWO='<';
                else if(s1[5]==s2[5]) TWO='=';
                else TWO='>';
            }
            else TWO='=';
        }
        else TWO='>';
    }
    else TWO='>';
}

int main()
{
    int T_T;
    int cas=1;
    scanf("%d",&T_T);
    while(T_T--)
    {
        scanf("%s%s",s1,s2);
        if(strlen(s1)<6) s1[5]='A';
        if(strlen(s2)<6) s2[5]='A';
        BJ1();BJ2();
        printf("Case %d: %c %c\n",cas++,ONE,TWO);
    }
    return 0;
}