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>
#pragma comment(linker, "/STACK:64000000")
using std::cerr;
using std::cin;
using std::cos;
using std::cout;
using std::endl;
using std::make_pair;
using std::map;
using std::max;
using std::min;
using std::next_permutation;
using std::pair;
using std::priority_queue;
using std::queue;
using std::random_shuffle;
using std::set;
using std::sin;
using std::sort;
using std::sqrt;
using std::stack;
using std::string;
using std::stringstream;
using std::swap;
using std::vector;
const long double PI = 3.14159265358979323846;
vector<int> used;
vector<vector<int> > en;
int main() {
  int k, b, n;
  cin >> k >> b >> n;
  vector<int> data(n);
  long long res = 0;
  for (int i = 0; i < n; ++i) {
    cin >> data[i];
  }
  long long z = 0;
  long long res0 = 0;
  for (int i = 0; i < n; ++i)
    if (data[i] == 0)
      ++z;
    else {
      res0 += z * (z + 1) / 2;
      z = 0;
    }
  res0 += z * (z + 1) / 2;
  if (b == 0) {
    cout << res0 << endl;
    return 0;
  }
  if (b == k - 1) b = 0;
  vector<int> sum(n + 1);
  map<int, int> num;
  num[0] = 1;
  sum[0] = 0;
  for (int i = 1; i <= n; ++i) {
    sum[i] = (sum[i - 1] + data[i - 1]) % (k - 1);
    int x = (sum[i] - b + k - 1) % (k - 1);
    res += num[x];
    if (num.find(sum[i]) == num.end())
      num[sum[i]] = 1;
    else
      ++num[sum[i]];
  }
  if (b == 0) res -= res0;
  cout << res << endl;
  return 0;
}
