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.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Locale;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;


public class Solution implements Runnable {
	
	BufferedReader in;
	PrintWriter out;
	StringTokenizer st;
	
	String nextToken() throws Exception {
		while (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(in.readLine());
		}
		return st.nextToken();
	}
	
	int nextInt() throws Exception {
		return Integer.parseInt(nextToken());
	}
	
	long nextLong() throws Exception {
		return Long.parseLong(nextToken());
	}
	
	void solve() throws Exception {
		int k = nextInt(), b = nextInt(), n = nextInt();
		int[] a = new int[n];
		int[] s = new int[n + 1];
		TreeMap<Integer, Integer> tm = new TreeMap<Integer, Integer>();
		tm.put(0, 0);
		for (int i = 0; i < n; i++) {
			s[i + 1] = (s[i] + (a[i] = nextInt())) % (k - 1);
			tm.put(s[i + 1], 0);
		}
		long ans = 0;
		for (int i = n; i >= 0; i--) {
			if (tm.containsKey((b + s[i]) % (k - 1))) ans += tm.get((b + s[i]) % (k - 1));
			tm.put(s[i], tm.get(s[i]) + 1);
		}
		if (b % (k - 1) == 0) {
			long t = 0;
			int bb = -1;
			for (int i = 0; i < n; i++) {
				if (a[i] != 0) {
					if (bb != -1) {
						t += (long)(i - bb) * (i - bb + 1) / 2;
						bb = -1;
					}
				} else {
					if (bb == -1) bb = i;
				}
			}
			if (bb != -1) t += (long)(n - bb) * (n - bb + 1) / 2;
			if (b == 0) ans = t; else ans -= t;
		}
		out.println(ans);
	}
	
	public void run() {
		try {
			Locale.setDefault(Locale.UK);
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			solve();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		} finally {
			out.close();
		}
	}

	public static void main(String[] args) {
		new Thread(null, new Solution(), "1", 1 << 28).start();
	}

}