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: Wrong Answer)

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

//Find the Multiples
public class Main{

	Map<Integer, Integer> cnt;
	void inc(int m){
		if(cnt.containsKey(m))cnt.put(m, cnt.get(m)+1);
		else cnt.put(m, 1);
	}
	
	void run(){
		Scanner sc = new Scanner(System.in);
		for(;;){
			int n = sc.nextInt(), S = sc.nextInt(), W = sc.nextInt(), Q = sc.nextInt();
			if((n|S|W|Q)==0)break;
			int[] a = new int[n];
			int g = S;
			for(int i=0;i<n;i++){
				a[i] = (g/7) % 10;
				if((g&1) == 0)g>>=1;
				else g = (g>>1) ^ W;
			}
			int res = 0;
			if(Q==2){
				int e = 0;
				for(int i=n-1;i>=0;i--){
					if((a[i]&1)==0)e++;
					if(a[i]!=0)res+=e;
				}
			}
			else if(Q==5){
				int f = 0;
				for(int i=n-1;i>=0;i--){
					if(a[i]==0||a[i]==5)f++;
					if(a[i]!=0)res+=f;
				}
			}
			else{
				cnt = new HashMap<Integer, Integer>();
				inc(0);
				int ten = 1, m = a[n-1]%Q;
				inc(m);
				if(m==0)res++;
				for(int i=n-2;i>=0;i--){
					ten = (ten*10)%Q;
					m = (a[i]*ten+m)%Q;
					if(a[i]!=0 && cnt.containsKey(m))res+=cnt.get(m);
					inc(m);
				}
			}
			System.out.println(res);
		}
	}

	public static void main(String[] args) {
		new Main().run();
	}
}