GOV02 - Good

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The problem is very easy.

You are given n terms ( the terms can be integer as well as floating ). Let S = sum of all the n terms. A number k (x<=k<=y) is said to be good if S is divisible by it.

Input

Input begins with an integer t.

Then t test cases follow.

For each test case, three numbers n,x,y are given. Then n terms of the sequence follow.

(REMEMBER: all the input is on a single line)

Output

For each test case,

You have to output a single integer m.

where, m = (sum of all even good numbers) - ( number of all odd good numbers)

(REMEMBER: all the output should be on a single line)

The numbers should be separated by spaces.

Example

Input:
2 3 1 2 1 2 3 4 1 10.5 -1 4.5 4.5 4

Output:
1 10

Scoring:

Your task is to minimise the source code length.

The less your fingers work, more you gain.

Remember:

Only python 2.7 is permitted. Sorry in advance, I will not allow any other language.

Source limit is tight. So be careful.

Constraints:

1<=t<=10

1<=n<=100

-100<=any term of sequence<=100

0<x<=y<=100

 


hide comments
Mitch Schwartz: 2015-02-16 14:49:29

@]V[AgN3To: I think the given cases are sufficient. If your code is mostly correct but fails for some tricky cases, I don't know what they would be; my initial approach (once I understood what the author wanted) was straightforward and passed.

Aditya Pande: 2015-02-16 14:49:29

NZEC after running(2)...?
shouldn't raw_input() and split() work ?

Last edit: 2013-01-16 02:14:54
Tushar Makkar (Retired): 2015-02-16 14:49:29

Can somebody explain me How the answer of 2nd case is 10 ?
@Mitch and @Govind : Can u give an example which satisfies all the constraints ?

Aditya Pande: 2015-02-16 14:49:29

why am i getting NZEC....?? please help....
i hope to have done it right...

please help. it has been hard enough to get it into 200 bytes
EDIT:
>>>NZEC again and again...?

Last edit: 2013-01-11 11:42:17

Added by:Govind Lahoti
Date:2013-01-07
Time limit:0.5s
Source limit:200B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:PYTHON
Resource:self