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.*;

public class Main{
	
	private void doit(){
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()){
			int n = sc.nextInt();
			int s = sc.nextInt();
			int w = sc.nextInt();
			int q = sc.nextInt();
			if((n|s|w|q) == 0) break;
			char [] a = new char[n];
			for(int i = 0, g = s; i < n; i++){
				a[i] = (char) (((g / 7) % 10) + '0');
				if(g % 2 == 0){
					g = (g / 2);
				}
				else{
					g = (g / 2) ^ w;
				}
			}
			int ans = 0;
			if(q == 2 || q == 5){
				for(int i = 0, count = 0; i < n; i++){
					if( (int)(a[i] - '0') % q == 0){
						if((int)(a[i] - '0') != 0){
							count++;
						}
						ans += count;
					}
				}
			}
			else{
				HashMap<Long, Integer> freq = new HashMap<Long, Integer>();
				freq.put(0L, 1);
				long digitUp = 1, now = 0;
				for(int i = n - 1; i >= 0; i--){
					now = ((long)(a[i] - '0') * digitUp + now) % q;
					digitUp = (digitUp * 10) % q;
					if(a[i] != '0' && freq.containsKey(now)){
						ans += freq.get(now);
					}
					
					if(freq.containsKey(now)){
						freq.put(now, freq.get(now) + 1);
					}
					else{
						freq.put(now, 1);
					}
				}
			}
			System.out.println(ans);
		}
	}

	private void debug(Object... o) {
		System.out.println("debug = " + Arrays.deepToString(o));
	}

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