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.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.HashMap;
import java.util.Map;

public class E216 {

    static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out = new PrintWriter(System.out);
    
    static int nextInt() throws Exception {
        in.nextToken();
        return (int)in.nval;
    }
    
    static String nextString() throws Exception {
        in.nextToken();
        return in.sval;
    }

    public static void main(String[] args) throws Exception {
        int k = nextInt(), b = nextInt(), n = nextInt();
        long ans = 0;
        if (b == 0) {
            long cnt = 0;
            for (int i = 0; i < n; i++) {
                int a = nextInt();
                if (a == 0) cnt++;
                else {
                    ans += cnt*(cnt+1)/2;
                    cnt = 0;
                }
            }
            ans += cnt*(cnt+1)/2;
        }
        else {
            Map<Integer, Integer> map = new HashMap<Integer, Integer>();
            map.put(0, 1);
            int x = 0, cnt = 0;
            for (int i = 0; i < n; i++) {
                int a = nextInt();
                if (a == 0) cnt++;
                else cnt = 0;
                x = (int)((1L*x*k + a)%(k-1));
                int t = (x - b + k - 1)%(k-1);
                ans += map.containsKey(t) ? map.get(t) : 0;
                if (b == k-1) ans -= cnt;
                if (!map.containsKey(x)) map.put(x, 1);
                else map.put(x, map.get(x)+1);
            }
        }
        out.println(ans);
        out.flush();
    }

}
