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.Map;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.TreeMap;
import java.util.InputMismatchException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;

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

class TaskE {
    long ansFor0(int[] a){
        long ans = 0;
        int cur = 0;
        for(int c: a){
            if(c == 0){
                ++cur;
                ans += cur;
            }
            else
                cur = 0;
        }
        return ans;
    }
	public void solve(int testNumber, Reader in, PrintWriter out) {
        int k = in.nextInt();
        int b = in.nextInt();
        int n = in.nextInt();
        int[] a = in.nextIntArray(n);

        long ans = 0;
        if(b == 0){

            ans = ansFor0(a);
        }
        else{
            long toSubstruct = 0;
            if(b == k - 1){
                toSubstruct = ansFor0(a);
                b = 0;
            }
            ans -= toSubstruct;
            Map<Integer, Integer> map = new TreeMap<Integer, Integer>();

            map.put(0, 1);
            int cur = 0;
            for(int c: a){
                cur += c;
                cur %= k - 1;
                if(map.containsKey((cur - b + k - 1) % (k - 1)))
                    ans += map.get((cur - b + k - 1) % (k - 1));

                if(map.containsKey(cur))
                    map.put(cur, map.get(cur) + 1);
                else
                    map.put(cur, 1);
            }

        }

        out.print(ans);
	}
}

class Reader {
    private BufferedReader reader;
    private StringTokenizer tokenizer;

    public Reader(BufferedReader reader) {
        this.reader = reader;
    }

    public Reader(InputStream stream) {
        this(new BufferedReader(new InputStreamReader(stream)));
    }

    public String nextString() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(readLine());
        }
        return tokenizer.nextToken();
    }

    public int nextInt() {
        return Integer.parseInt(nextString());
    }

    public int[] nextIntArray(int size) {
        int[] array = new int[size];
        for (int i = 0; i < size; ++i) {
            array[i] = nextInt();
        }
        return array;
    }

    private String readLine() {
        try {
            return reader.readLine();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }
}

