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 Round133Div2A {
	public static void main(String[] args) throws IOException {
		rd = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(System.out);
		st = new StringTokenizer(rd.readLine());
		int k = Integer.parseInt(st.nextToken()), b = Integer.parseInt(st.nextToken()), n =  Integer.parseInt(st.nextToken());
		int[] A = new int[n];
		st = new StringTokenizer(rd.readLine());
		int M = (k-1);
		int[] S = new int[n];
		for(int i=0; i<n; i++){
			A[i] = Integer.parseInt(st.nextToken());
		}
		if(b==0){
			long ans = 0;
			for(int i=0; i<n; i++){
				if(A[i]!=0) continue;
				int max = i;
				for(int j=i; j<n; j++) if(A[j]==0) max = j; else break;
				long len = max-i+1;
				i = max;
				ans += (len*(len+1))>>1;
			}
			System.out.println(ans);
			return;
		}
		if(k==2){
			long ans = n;
			ans = (ans*(ans+1))>>1;
			for(int i=0; i<n; i++){
				if(A[i]!=0) continue;
				int max = i;
				for(int j=i; j<n; j++) if(A[j]==0) max = j; else break;
				long len = max-i+1;
				i = max;
				ans -= (len*(len+1))>>1;
			}
			System.out.println(ans);
			return;
		}
		S[0] = A[0]%M;
		for(int i=1; i<n; i++) S[i] = (S[i-1] + A[i])%M;
		HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
		for(int i=0; i<n; i++){
			int cur = S[i];
			if(!map.containsKey(cur)) map.put(cur, 1);
			else map.put(cur, map.get(cur)+1);
		}
		long ans = 0;
		for(int i=0; i<=n; i++){
			int cur = (i==0)? 0: S[i-1];
			int desired = (cur+b)%M;
			int app = !map.containsKey(desired)? 0: (desired!=cur? map.get(desired): map.get(desired)-1);
			if(i==0) app = !map.containsKey(desired)? 0: map.get(desired);
			if(map.containsKey(desired)) ans += app;
			if(i!=0){
				int num = S[i-1];
				if(map.get(num)==1) map.remove(num);
				else map.put(num, map.get(num)-1);
			}
		}
		if(b!=k-1)
			System.out.println(ans);
		else{
			for(int i=0; i<n; i++){
				if(A[i]!=0) continue;
				int max = i;
				for(int j=i; j<n; j++) if(A[j]==0) max = j; else break;
				long len = max-i+1;
				i = max;
				ans -= (len*(len+1))>>1;
			}
			System.out.println(ans);
		}
	}
	
	
	
	
	
	
	static BufferedReader rd;
	
	static PrintWriter pw;

	static StringTokenizer st;
	
}
