LASTDIG  The last digit
Nestor was doing the work of his math class about three days but he is tired of make operations a lot and he should deliver his task tomorrow. His math’s teacher gives him two numbers a and b. The problem consist of finding the last digit of the potency of base a and index b. Help Nestor with his problem. You are given two integer numbers: the base a (0 <= a <= 20) and the index b (0 <= b <= 2,147,483,000), a and b both are not 0. You have to find the last digit of a^{b}.
Input
The first line of input contains an integer t, the number of test cases (t <= 30). t test cases follow. For each test case will appear a and b separated by space.
Output
For each test case output an integer per line representing the result.
Example
Input: 2 3 10 6 2
Output: 9 6
hide comments
shantanu_25:
20200917 16:57:08
I think here "a and b both are not 0" means both are not 0 at the same time. 

rudra_hari007:
20200915 07:33:03
Can anyone tell why normal solution is not working? 

jinks:
20200905 16:55:26
I'm new to SPOJ, I've submitted my solution but it crossed the memory limit of 70 bytes. How did you guys manage to solve it in less than 700 bytes? 

thepankj:
20200817 09:17:46
Misleading statement 'a and b both are not 0'.


hex1729:
20200814 00:24:44
No exponentiation stuff required if you use your brains in last digit patterns of powers :)


kishlay1105:
20200726 10:25:54
calculate (a^b)mod 10 using binary exponentiation . 

dungeon_mr123:
20200725 17:50:47
i cannot what is mistake in code please help


gokulan_cv:
20200708 10:12:01
AC After 14 Times of failed submission. Python is horrible for CP. I think I should start learning c++.


acktron:
20200630 12:10:36
Solved in one go using modular exponentiation four : (a ^ n) % 10 = ( ( (a % 10) ^ n); 

br0ken_c0der:
20200629 07:44:07
hint: Apply fast modular exponentiation using bitmasking. 
Added by:  Jose Daniel Rodriguez Morales 
Date:  20081201 
Time limit:  1s 
Source limit:  700B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: GOSU 
Resource:  Own 