Miaomiao's Function

Accepts: 6
Submissions: 113
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
Problem Description
Firstly , Miaomiao define two functions f(x) , g(x): [img]../../../data/images/C528-1002-2.jpg[/img] (K is the smallest integer which satisfied x + 9 * K > 0) if [img]../../../data/images/C528-1002-3.jpg[/img] e.g. g(178) = 1 - 7 + 8 = 2 , g(1) = 1 , g(1234) = 1 - 2 + 3 - 4 = -2; For example f(20140810) = f(2 + 0 + 1 + 4 + 0 + 8 + 1 + 0) = f(16) = f(1 + 6) = f(7) = 7 Then , you are given two integers L , R( L <= R) . Answer is defined as [img]../../../data/images/C528-1002-1.jpg[/img]. Please caculate (Answer Mod f(Answer) + f(Answer)) Mod f(Answer). Pay attantion ! If f(Answer) = 0 , please output "Error!"
Input
There are many test cases. In the first line there is a number T ( T <= 50 ) shows the number of test cases. For each test cases there are two numbers L ,R ( 0 <= L,R <= 10^100 ). For your Information , L , R don't have leading zeros.
Output
For each opeartion , output the result.
Sample Input
2
0 0
0 21
Sample Output
Error!
1
Hint
The Answer is 0 and 13. So the result is Error! and 13 Mod (1 + 3) = 1