Lucky Days

All submissions for this problem are available.
Chef Ciel defined a sequence S as follows:
S[1] = A
S[2] = B
S[i] = (X*S[i1] + Y*S[i2] + Z) mod P, for i >= 3
Ciel considers C is a lucky number, and the ith day is a lucky day if and only if S[i] = C.
Ciel's restaurant may have special events in a lucky day.
By the way, your work is calculating the numbers of lucky days in intervals.
That is, for each Q intervals [L[i], R[i]],
you should calculate the number of the integers k which satisfy L[i] <= k <= R[i] and S[k] = C.
Input
The first line contains an integer T, the number of test cases.
Then T test cases follow.
The first line for each test case has 8 integers A, B, X, Y, Z, P, C and Q.
The next Q lines have 2 integers L[i] and R[i].
Output
For each interval, print the number of lucky days in the interval.
Constraints
1 <= T <= 2
2 <= P <= 10007
P is a prime.
0 <= A, B, X, Y, Z, C < P
1 <= Q <= 20000 (2*10^4)
1 <= L[i] <= R[i] <= 1000000000000000000 (10^18)
Sample Input
2 1 1 1 1 0 2 0 6 1 1 2 2 3 3 4 4 5 5 6 6 1 2 4 5 3 17 4 3 5 8 5 58 58 5858
Sample Output
0 0 1 0 0 1 0 4 362
Output details
In the first case:
S[1] = A = 1
S[2] = B = 1
S[3] = (S[2] + S[1]) mod 2 = (1 + 1) mod 2 = 0
S[4] = (S[3] + S[2]) mod 2 = (0 + 1) mod 2 = 1
S[5] = (S[4] + S[3]) mod 2 = (1 + 0) mod 2 = 1
S[6] = (S[5] + S[4]) mod 2 = (1 + 1) mod 2 = 0
S[7] = (S[6] + S[5]) mod 2 = (0 + 1) mod 2 = 1
Author:  laycurse 
Tester:  chmel_tolstiy 
Editorial  http://discuss.codechef.com/problems/LUCKYDAY 
Tags  hard laycurse nov11 
Date Added:  1102011 
Time Limit:  5 sec 
Source Limit:  50000 Bytes 
Languages: 
Comments
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SUCCESSFUL SUBMISSIONS
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Should we consider number P
Oops. Found it in the text.
For: 1 2 4 5 3 17 4 3 5 8 5
sorry my mistake!!
3 correct "JAVA" submissions,
My code exits with runtime
do we have to print the
@admin i am getting a runtime
Can X,Y and Z be negative
@chandransuraj if they are
@chandansuraj  carefully
Thanks a lot guys...my bad
am I missing something or the
theicebreaker: The limits are
hi guys. i'm printing some