SKIT's Chemistry Department found some liquid containing a precious chemical which can be separated out of the liquid using centrifugation.
This liquid need to be centrifuged for 5 minutes continuously after that some another facilitating chemical is mixed in the liquid and it is again centrifuged for 5 minutes. 
It is not necessary to perform the centrifugation of same tube successively.

You are given the maximum capacity of Centrifuge i.e. the number of tubes it can hold in a round of centrifugation as M.

You are also given the total number of tubes to be centrifuged as N.

You are given the task to find the minimum time in which all the tubes can be centrifuged completely.

Input:

First line of input contains an integer T denoting number of test cases.
Each test case contains two space separated integers M and N.

Output:

Print the minimum time according to test case.

Constraints:
1 ≤ T ≤ 40
1 ≤ M ≤ 1000
0 ≤ N ≤ 1000

SAMPLE INPUT
2
2 3
3 4

SAMPLE OUTPUT
15
15

Explanation

Test Case #1:

Let's say A,B,C are the three tubes of liquid.
Firstly A and B are centrifuged for 5 minutes then after mixing chemicals in B, B and C can be centrifuged for 5 minutes, hence after 10 minutes B is complete and A, C are now centrifuged for 5 minutes.
Hence total time taken for complete centrifugation is : 5 + 5 + 5  = 15
for _ in xrange(input()):
	Cap, Max = map(int,raw_input().split());
	if Max <= Cap and Max != 0:
		print 10;
	else:
		x = (Max * 2.0) / Cap * 1.0;
		if x > int(x):
			print (int(x)+1)*5;
		else:
			print int(x)*5;