You know that the Martians use a number system with base k. Digit b (0 ≤ b < k) is considered lucky, as the first contact between the Martians and the Earthlings occurred in year b (by Martian chronology).

A digital root d(x) of number x is a number that consists of a single digit, resulting after cascading summing of all digits of number x. Word "cascading" means that if the first summing gives us a number that consists of several digits, then we sum up all digits again, and again, until we get a one digit number.

For example, d(35047) = d((3 + 5 + 0 + 4)7) = d(157) = d((1 + 5)7) = d(67) = 67. In this sample the calculations are performed in the 7-base notation.

If a number's digital root equals b, the Martians also call this number lucky.

You have string s, which consists of n digits in the k-base notation system. Your task is to find, how many distinct substrings of the given string are lucky numbers. Leading zeroes are permitted in the numbers.

Note that substring s[i... j] of the string s = a1a2... an (1 ≤ i ≤ j ≤ n) is the string aiai + 1... aj. Two substrings s[i1... j1] and s[i2... j2] of the string s are different if either i1 ≠ i2 or j1 ≠ j2.

Input

The first line contains three integers k, b and n (2 ≤ k ≤ 109, 0 ≤ b < k, 1 ≤ n ≤ 105).

The second line contains string s as a sequence of n integers, representing digits in the k-base notation: the i-th integer equals ai (0 ≤ ai < k) — the i-th digit of string s. The numbers in the lines are space-separated.

Output

Print a single integer — the number of substrings that are lucky numbers.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

10 5 6
3 2 0 5 6 1


Output

5

Input

7 6 4
3 5 0 4


Output

1

Input

257 0 3
0 0 256


Output

3

Note

In the first sample the following substrings have the sought digital root: s[1... 2] = "3 2", s[1... 3] = "3 2 0", s[3... 4] = "0 5", s[4... 4] = "5" and s[2... 6] = "2 0 5 6 1".
import java.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.HashMap;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author George Marcus
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskE solver = new TaskE();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskE {
	public void solve(int testNumber, Scanner in, PrintWriter out) {
        int K = in.nextInt();
        int B = in.nextInt();
        int N = in.nextInt();
        int[] s = new int[N];
        for(int i = 0; i < N; i++)
            s[i] = in.nextInt();

        long zeroSubsequences = 0;
        int crtZeros = 0;

        for(int i = 0; i < N; i++) {
            if(s[i] == 0)
                crtZeros++;
            else
                crtZeros = 0;
            zeroSubsequences += crtZeros;
        }

        if(B == 0) {
            out.print(zeroSubsequences);
            return;
        }

        HashMap<Integer, Integer> dp = new HashMap<Integer, Integer>();

        dp.put(0, 1);

        long answer = 0;
        int crtRoot = 0;
        for(int i = 0; i < N; i++) {
            crtRoot = (crtRoot + s[i]) % (K - 1);
            int complement = (crtRoot - B + K - 1) % (K - 1);

            if(dp.containsKey(complement))
                answer += dp.get(complement);

            if(dp.containsKey(crtRoot))
                dp.put(crtRoot, dp.get(crtRoot) + 1);
            else
                dp.put(crtRoot, 1);
        }

        if(B == K - 1)
            answer -= zeroSubsequences;

        out.print(answer);
	}
}

