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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1963    Accepted Submission(s): 575


Problem Description
We all know a+b=c,but now there is a new rule in equations:
if a>b,then it's equal to [>c];
if a=b,then it's equal to [=c];
if a<b,then it's equal to [<c];
a sample:

1+1=[=2]
2+0=[>2]
0+2=[<2]
1+2=[<3]

and [>c] > [=c] > [<c].
For every [n],
[n+1] > [n] >[n-1].
 

Input
The input consists of T cases.The first line of the input contains only positive integer T (0<T<=100).Then follows the cases.Each case begins with a line containing exactly one integer N (0<N<=10000).Then there are N lines, each with two integers a and b (0<a,b<2^31-1).
 

Output
For each case, calculate each value of a+b(as a+b=[?]) and output a+b=[?] in descending order sorted by [?]. If there exists some pair of a,b whose order can not be determined by the rules above, output them with the order of the input. A blank line should be printed after each case.
 

Sample Input
2 5 1 5 5 1 3 3 4 5 5 6 5 1 6 6 1 2 4 4 2 3 3
 

Sample Output
5+6=[<11] 4+5=[<9] 5+1=[>6] 3+3=[=6] 1+5=[<6] 6+1=[>7] 1+6=[<7] 4+2=[>6] 3+3=[=6] 2+4=[<6]
 

Source
 

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