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".
#include <bits/stdc++.h>
using namespace std;
long long nn(long long x, long long m) {
  if (x == 0) return 0;
  x = (x % m + m) % m;
  if (x == 0) return m;
  return x;
}
long long mm(long long x, long long m) { return (x % m + m) % m; }
int main() {
  long long n, k, b;
  long long ps = 0, ans = 0, cnt[111111], ls = -1;
  map<long long, long long> MP;
  cin >> k >> b >> n;
  MP[0] = 1;
  for (int i = (0); i < (n); i++) {
    long long x;
    cin >> x;
    ps += x;
    if (x == 0) {
      if (ls == -1) ls = i;
    } else {
      ls = -1;
    }
    if (b == 0) {
      cnt[i] = ls == -1 ? 0 : i - ls + 1;
    } else {
      if (ls > 0)
        cnt[i] = cnt[ls - 1];
      else if (ls == 0)
        cnt[i] = 0;
      else {
        long long y = mm(ps - b, k - 1);
        if (MP.find(y) == MP.end())
          cnt[i] = 0;
        else
          cnt[i] = MP[y];
      }
    }
    if (MP.find(ps % (k - 1)) != MP.end()) {
      MP[ps % (k - 1)] = MP[ps % (k - 1)] + 1;
    } else {
      MP[ps % (k - 1)] = 1;
    }
    ans += cnt[i];
  }
  cout << ans << endl;
  return 0;
}
