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 javax.jws.Oneway;
import java.io.*;
import java.math.BigInteger;
import static java.math.BigInteger.*;

import java.security.PublicKey;
import java.util.*;

public class Solution{


    int k,b,n;
    int[]a;
    long countZero()
    {
        long res=0;
        int i=0;
        while (i<n)
        {
            if(a[i]==0)
            {
                int cur=0;
                while (i<n && a[i]==0)
                {
                    i+=1;
                    cur+=1;
                }
                res+=(long)(cur)*(cur-1)/2+cur;
            }
            else
                i+=1;
        }
        return res;
    }

    long simple()
    {


    long res=0;
    HashMap<Integer,Integer>sum=new HashMap<Integer, Integer>();
    sum.put(0,1);
    int cur=0;
    for (int i=0;i<n;i++)
    {
        cur+=a[i];
        cur%=(k-1);
        int v=(cur-b+k-1)%(k-1);
         if(sum.containsKey(v))
              res+=sum.get(v);

        if(!sum.containsKey(cur))
            sum.put(cur,1);
        else
            sum.put(cur,sum.get(cur)+1);
    }
    return res  ;
    }
    void solve()throws Exception
    {

         k=nextInt();
        b=nextInt();
        n=nextInt();

        a=new int[n];
        for(int i=0;i<n;i++)
            a[i]=nextInt();



        long res=0;
        if(b==0)
            res=countZero();
        else if(b!=k-1)
            res=simple();
        else
        {
            b=0;
            res=simple();
            res-=countZero();
        }
        System.out.println(res);


    }
    ////////////
    BufferedReader reader;
    PrintWriter writer;
    StringTokenizer stk;
    void run()throws Exception
    {

        reader=new BufferedReader(new InputStreamReader(System.in));
        // reader=new BufferedReader(new FileReader("input.txt"));
        stk=null;
        writer=new PrintWriter(new PrintWriter(System.out));
        //writer=new PrintWriter(new FileWriter("output.txt"));
        solve();
        reader.close();
        writer.close();
    }
    int nextInt()throws Exception
    {
        return Integer.parseInt(nextToken());
    }

    long nextLong()throws Exception
    {
        return Long.parseLong(nextToken());

    }
    double nextDouble()throws Exception
    {
        return Double.parseDouble(nextToken());


    }

    String nextString()throws Exception
    {
        return nextToken();
    }
    String nextLine()throws Exception
    {
        return reader.readLine();
    }
    String nextToken()throws Exception
    {
        if(stk==null || !stk.hasMoreTokens())
        {
            stk=new StringTokenizer(nextLine());
            return nextToken();

        }
        return stk.nextToken();
    }

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








}
