Problem:
Problem F: Find the Multiples

You are given a sequence a0a1...aN-1 digits and a prime number Q. For each i ≤ j with ai ≠ 0, the subsequence aiai+1...aj can be read as a decimal representation of a positive integer. Subsequences with leading zeros are not considered. Your task is to count the number of pairs (i, j) such that the corresponding subsequence is a multiple of Q.

Input

The input consists of at most 50 datasets. Each dataset is represented by a line containing four integers N, S, W, and Q, separated by spaces, where 1 ≤ N ≤ 105, 1 ≤ S ≤ 109, 1 ≤ W ≤ 109, and Q is a prime number less than 108. The sequence a0...aN-1 of length N is generated by the following code, in which ai is written as a[i].

int g = S;
for(int i=0; i<N; i++) {
a[i] = (g/7) % 10;
if( g%2 == 0 ) { g = (g/2); }
else { g = (g/2) ^ W; }
}

Note: the operators /, %, and ^ are the integer division, the modulo, and the bitwise exclusiveor, respectively. The above code is meant to be a random number generator. The intended solution does not rely on the way how the sequence is generated.

The end of the input is indicated by a line containing four zeros separated by spaces.

Output

For each dataset, output the answer in a line. You may assume that the answer is less than 230.

Sample Input

3 32 64 7
4 35 89 5
5 555 442 3
5 777 465 11
100000 666 701622763 65537
0 0 0 0

Output for the Sample Input

2
4
6
3
68530

In the first dataset, the sequence is 421. We can find two multiples of Q = 7, namely, 42 and 21.

In the second dataset, the sequence is 5052, from which we can find 5, 50, 505, and 5 being the multiples of Q = 5. Notice that we don't count 0 or 05 since they are not a valid representation of positive integers. Also notice that we count 5 twice, because it occurs twice in different positions.

In the third and fourth datasets, the sequences are 95073 and 12221, respectively.

Solution in JAVA (judge verdict: Memory Limit Exceeded)

import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (true) {
			int N = sc.nextInt();
			int S = sc.nextInt();
			int W = sc.nextInt();
			int Q = sc.nextInt();
			if (N == 0 && S == 0 && W == 0 && Q == 0)
				break;
			int g = S;
			int[] a = new int[N];
			for (int i = 0; i < N; ++i) {
				a[N - 1 - i] = (g / 7) % 10;
				if (g % 2 == 0) {
					g = (g / 2);
				} else {
					g = (g / 2) ^ W;
				}
			}

			int[] seq = new int[N];
			long pow10 = 1;
			int zeropos = 0;
			int[] count = new int[Q];
			count[0] = 1;
			int ans = 0;
			for (int i = 0; i < N; ++i) {
				if (Q != 2 && Q != 5) {
					seq[i] = (int) (((i > 0 ? seq[i - 1] : 0) + pow10 * a[i]) % Q);
					if (a[i] != 0)
						ans += count[seq[i]];
					++count[seq[i]];
					pow10 = pow10 * 10 % Q;
				} else {
					if (a[i] % Q == 0) {
						++zeropos;
					}
					if (a[i] != 0) {
						ans += zeropos;
					}
				}
			}
			System.out.println(ans);
			System.gc();
		}
	}
}