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 E {
    public static void main(String[] args) {
        try {
            new E().solve();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    int k, b, n;
    int[] a;
    void solve() throws IOException {
        // BufferedReader in = new BufferedReader(new
        // InputStreamReader(System.in));
        Scanner sc = new Scanner(System.in);
        
        k = sc.nextInt();
        b = sc.nextInt();
        n = sc.nextInt();
        a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        if (b == 0) {
            long ret = 0;
            int st = 0;
            while (true) {
                while (st < n && a[st] != 0) {
                    st++;
                }
                int end = st + 1;
                while (end < n && a[end] == 0) {
                    end++;
                }
                if (end <= n) {
                    int nn = end - st;
                    ret += (nn * (long) (nn + 1)) / 2;
                    st = end;
                } else {
                    break;
                }
            }
            System.out.println(ret);
            
        } else {
            System.out.println(solve2(0, n));
        }
    }
    HashMap<Long, Integer> map = new HashMap<Long, Integer>();
    long solve2(int st, int end) {
        int nn = end - st;
        if (nn < 20) {
            long ret = 0;
            for (int from = st; from < end; from++) {
                long sum = 0;
                for (int i = from; i < end; i++) {
                    sum += a[i];
                    if (b == k - 1 ? sum != 0 && sum % (k - 1) == 0
                            : sum % (k - 1) == b) {
                        ret++;
                    }
                }
            }
            return ret;
            
        } else {
            int m = st + nn / 2;
            long ret = 0;
            ret += solve2(st, m) + solve2(m, end);
            
            map.clear();
            {
                long Rsum = 0;
                for (int i = m; i < end; i++) {
                    Rsum += a[i];
                    if (Rsum != 0 && Rsum % (k - 1) == 0) {
                        Rsum = k - 1;
                    } else {
                        Rsum %= k - 1;
                    }
                    if (map.containsKey(Rsum)) {
                        map.put(Rsum, map.get(Rsum) + 1);
                    } else {
                        map.put(Rsum, 1);
                    }
                }
            }
            
            long Lsum = 0;
            for (int i = m - 1; i >= st; i--) {
                Lsum += a[i];
                if (Lsum != 0 && Lsum % (k - 1) == 0) {
                    Lsum = k - 1;
                } else {
                    Lsum %= k - 1;
                }
                
                long r = (b - Lsum + k - 1) % (k - 1);// 0~k-2
                if (r == 0) {
                    if (Lsum != 0) {
                        // 0ok
                        if (map.containsKey(r)) {
                            ret += map.get(r);
                        }
                    }
                    r = k - 1;
                }
                if (map.containsKey(r)) {
                    ret += map.get(r);
                }
            }
            return ret;
        }
    }
}

















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