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.*;
import java.util.*;

public class P216E
{
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        long k = sc.nextLong();
        long b = sc.nextLong();
        long n = sc.nextLong();
        long[] s = new long[(int)n];
        long[] c = new long[(int)n];
        for (int i = 0; i < n; i++)
        {
            s[i] = sc.nextLong();
        }
        if (k == 2 || b == 0)
        {
            solve2(b, s);
            return;
        }
        if (b == k - 1) b = 0;
        HashMap<Long, List<Integer>> map = new HashMap<Long, List<Integer>>();
        c[0] = s[0] % (k-1);
        List<Integer> alist = new ArrayList<Integer>();
        alist.add(0);
        map.put(c[0], alist);
        for (int i = 1; i < n; i++)
        {
            c[i] = (c[i-1] + s[i]) % (k-1);
            if (map.containsKey(c[i]))
            {
                map.get(c[i]).add(i);
            }
            else
            {
                List<Integer> list = new ArrayList<Integer>();
                list.add(i);
                map.put(c[i], list);
            }
        }
        long sum = 0;
        for (int i = 0; i < n; i++)
        {
            long val = i == 0 ? 0 : c[i-1];
            long tf = (b + val) % (k-1); 
            List<Integer> list = map.get(new Long(tf));
            if (list != null)
            {
                int pos = countge(list, i);
                sum += pos;
            }
        }
        if (b != 0)
        {
            System.out.println(sum);
        }
        else
        {
            System.out.println(sum - count0(s));
        }
        
    }
    
    private static int countge(List<Integer> list, int num)
    {
        if (num <= list.get(0)) return list.size();
        if (num > list.get(list.size() - 1)) return 0;
        int bot = 1, top = list.size() - 1;
        while (top > bot)
        {
            int mid = (bot + top) / 2;
            if (list.get(mid) < num)
            {
                bot = mid + 1;
            }
            else 
            {
                top = mid;
            }
        }
        return list.size() - bot;
    }
    
    private static long count0(long[] s)
    {
        long sum = 0;
        long start = -1;
        for (int i = 0; i < s.length; i++)
        {
            if (start == -1)
            {
                if (s[i] == 0)
                {
                    start = i;
                }
            }
            else
            {
                if (s[i] != 0)
                {
                    long len = i - start;
                    sum += (len * (len + 1)) / 2;
                    start = -1;
                }
            }
        }
        if (start != -1)
        {
            long len = s.length - start;
            sum += (len * (len + 1)) / 2;
        }
        return sum;
    }
    
    private static void solve2(long b, long[] s)
    {
        long sum = count0(s);
        long len = s.length;
        long tot = (len * (len + 1)) / (long)2;
        if (b == 0)
        {
            System.out.println(sum);
        }
        else
        {
            long ans = tot - sum;
            System.out.println(ans);
        }
    }
}
