#include <bits/stdc++.h>
using namespace std;
template <class T>
inline int size(const T& c) {
  return c.size();
}
long long T[100010];
long long S[100010];
int A[100010];
map<int, int> G;
map<long long, int> F;
int n, k, b;
long long res;
int main() {
  scanf("%d %d %d", &k, &b, &n);
  G[0]++;
  F[0]++;
  for (int i = (1), _b = (n); i <= _b; i++) {
    scanf("%d", &A[i]);
    S[i] = (S[i - 1] + A[i]) % (k - 1);
    if (b != 0) res = res + G[(S[i] - b + k - 1) % (k - 1)];
    G[S[i]]++;
    T[i] = T[i - 1] + A[i];
    if (b == 0)
      res = res + F[T[i]];
    else if (b == k - 1)
      res = res - F[T[i]];
    F[T[i]]++;
  }
  cout << res;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline int size(const T& c) {
  return c.size();
}
long long T[100010];
long long S[100010];
int A[100010];
map<int, int> G;
map<long long, int> F;
int n, k, b;
long long res;
int main() {
  scanf("%d %d %d", &k, &b, &n);
  G[0]++;
  F[0]++;
  for (int i = (1), _b = (n); i <= _b; i++) {
    scanf("%d", &A[i]);
    S[i] = (S[i - 1] + A[i]) % (k - 1);
    if (b != 0) res = res + G[(S[i] - b + k - 1) % (k - 1)];
    G[S[i]]++;
    T[i] = T[i - 1] + A[i];
    if (b == 0)
      res = res + F[T[i]];
    else if (b == k - 1)
      res = res - F[T[i]];
    F[T[i]]++;
  }
  cout << res;
}

```