SPOJ Problem Set (classical)
10966. Arithogeometric Series (AGS)
Problem code: AGS

Arithmatic and geometric Progressions are 2 of the well known progressions in maths.
Arithmatic progression(AP) is a set in which the difference between 2 numbers in constant. for eg, 1,3,5,7,9 .... In this series the difference between 2 numbers is 2.
Geometric progression(GP) is a set in which the ratio of 2 consecutive numbers is same. for eg, 1,2,4,8,16.... In this the ratio of the numbers is 2.
.....
What if there is a series in which we multiply a(n) by 'r' to get a(n+1) and then add 'd' to a(n+1) to get a(n+2)...
For eg .. lets say d=1 and r=2 and a(1) = 1..
series would be 1,2,4,5,10,11,22,23,46,47,94,95,190 ......
We add d to a(1) and then multiply a(2) with r and so on ....
Your task is, given 'a' , 'd' & 'r' to find the a(n) term .
sicne the numbers can be very large , you are required to print the numbers modulo 'mod'  mod will be supplied int the test case.
Input
first line of input will have number 't' indicating the number of test cases.
each of the test cases will have 2 lines
firts line will have 3 numbers 'a' ,'d' and 'r'
2nd line will have 2 numbers 'n' & 'mod'
a first term of the AGS
dthe difference element
r  the ratio element
n nth term required to be found
mod need to print the result modulo mod
Output
For each test case print "a(n)%mod" in a separate line.
Example
Input:
2
1 1 2
13 7
2 2 2
10 8
Output:
1
6
Description  for the first test case the series is 1,2,4,5,10,11,22,23,46,47,94,95,190..
13th term is 190 and 190%7 = 1
Note  the value of a , d , r , n & mod will be less than 10^8 and more than 0.
for every series 2nd term will be a+d and third term will be (a+d)*r .. and so on ..