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.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class Main {

    
    public static void main(String[] args) {
        
        InputScanner s = new InputScanner(System.in);
        
        int base = s.nextInt();
        int lucky = s.nextInt();
        int n = s.nextInt();
        int a[] = new int[n];
        for (int i = 0; i < a.length; i++) {
            a[i] = s.nextInt();
        }
        
        long zeroesAns = 0;
        int zeroes = 0;
        for (int i = 0; i < n; i++) {
            if(a[i] == 0)
                zeroes++;
            else zeroes = 0;
            zeroesAns += zeroes;
        }
        
        TreeMap<Integer, Integer> tree = new TreeMap<Integer, Integer>();
        
        long ans = 0;
        int sum = 0;
        tree.put(0, 1);
        for(int i: a){
            sum += i;
            if(sum >= base-1)
                sum -= base-1;
            int req = sum - lucky;
            if(req < 0)
                req += base - 1;
            
//          System.out.println(i+" "+req+ "   "+tree.get(req));
            
            Integer num = tree.get(req);
            if(num == null) num = 0;
            ans += num;
            
            Integer val = tree.get(sum);
            if(val == null) val = 0;
            tree.put(sum, val + 1);
            
        }
        
        if(lucky == 0){
            System.out.println(zeroesAns);
            return ;
        }
        if(lucky == base - 1){
            ans -= zeroesAns;
            System.out.println(ans);
            return ;
        }
        System.out.println(ans);
    }

}


class InputScanner {
    BufferedReader br;
    StringTokenizer strtok = null;

    public InputScanner(InputStream stream) {
        br = new BufferedReader(new InputStreamReader(stream));

    }

    public InputScanner(File file) throws FileNotFoundException {
        br = new BufferedReader(new FileReader(file));
    }

    public String next() {
        if (strtok == null || !strtok.hasMoreTokens()) {
            try {
                strtok = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return strtok.nextToken();
    }

    public int nextInt() {
        int i;
        try {
            i = Integer.parseInt(next());
        } catch (NumberFormatException e) {
            throw new RuntimeException(e);
        }
        return i;
    }

    public String nextLine() {
        if (strtok == null || !strtok.hasMoreTokens()) {
            try {
                strtok = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return strtok.nextToken("\n").trim();
    }

}
