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.util.HashMap;
import java.util.StringTokenizer;

public class PrE {

	public static long time;
	

	public static void main(String[] args) throws Exception {
		time = System.currentTimeMillis();
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		System.out.println(go(br));
		br.close();
		System.exit(0);
	}

	public static long go(BufferedReader br) throws Exception {
		StringTokenizer st = new StringTokenizer(br.readLine());
		int k = Integer.parseInt(st.nextToken()) - 1;
		int b = Integer.parseInt(st.nextToken());
		int n = Integer.parseInt(st.nextToken());
		st = new StringTokenizer(br.readLine());
		long num0 = 0, str0 = 0;
		long out = 0;
		int sum = 0;
		HashMap<Integer, Integer> num = new HashMap<Integer,Integer>(10000);
		num.put(0, 1);
		for(; n > 0; n--) {
			int cur = Integer.parseInt(st.nextToken());
			if(cur == 0) {
				str0 ++;
				num0 += str0;
			} else str0 = 0;
			sum += cur;
			sum %= k;
			Integer i = num.get((sum - b + k) % k);
			if(i == null) i = 0;
			out += i;
			i = num.get(sum);
			if(i == null) i = 0;
			num.put(sum, i + 1);
		}
		if(b == 0) return num0;
		if(b == k) return out - num0;
		return out;
	}

	public static void checkTime() {
		System.out.println(System.currentTimeMillis() - time);
		time = System.currentTimeMillis();
	}
}