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;
vector<vector<int> > a;
vector<bool> v;
int main() {
  int k, b, n;
  cin >> k >> b >> n;
  vector<int> v(n + 1);
  vector<std::pair<int, int> > s(n + 1);
  s[0] = make_pair(0, 0);
  long long int res = 0;
  int zeros = 0;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    if (v[i]) {
      res -= zeros * 1ll * (zeros + 1) / 2;
      zeros = 0;
    } else
      zeros++;
    s[i].first = (s[i - 1].first + v[i]) % (k - 1);
    s[i].second = i;
  }
  res -= zeros * 1ll * (zeros + 1) / 2;
  if (b == 0) {
    cout << -res;
    return 0;
  } else if (b != k - 1)
    res = 0;
  sort(s.begin(), s.end());
  for (int i = 1; i <= n; i++) {
    auto iter1 = lower_bound(s.begin(), s.end(),
                             make_pair((s[i].first - b + k - 1) % (k - 1), 0));
    auto iter2 = upper_bound(
        s.begin(), s.end(),
        make_pair((s[i].first - b + k - 1) % (k - 1), s[i].second - 1));
    res += iter2 - iter1;
  }
  cout << res;
}
