#include <bits/stdc++.h>
using namespace std;
long long ADD(long long a, long long b, int base) {
  base--;
  return (a + b + base) % base;
}
map<int, int> store;
int main() {
  store[0] = 1;
  long long base, b, n, a, csum = 0, len = 0, zeros = 0;
  long long ans = 0;
  cin >> base >> b >> n;
  for (int i = 0; i < n; i++) {
    cin >> a;
    if (a == 0)
      len++;
    else
      zeros += (long long)len * (len + 1) / 2, len = 0;
    csum = ADD(csum, a, base);
    a = ADD(csum, -b, base);
    ans += store[a];
    store[csum]++;
  }
  zeros += (long long)len * (len + 1) / 2;
  if (b == 0)
    cout << zeros << endl;
  else if (b == base - 1)
    cout << ans - zeros << endl;
  else
    cout << ans << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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;
long long ADD(long long a, long long b, int base) {
  base--;
  return (a + b + base) % base;
}
map<int, int> store;
int main() {
  store[0] = 1;
  long long base, b, n, a, csum = 0, len = 0, zeros = 0;
  long long ans = 0;
  cin >> base >> b >> n;
  for (int i = 0; i < n; i++) {
    cin >> a;
    if (a == 0)
      len++;
    else
      zeros += (long long)len * (len + 1) / 2, len = 0;
    csum = ADD(csum, a, base);
    a = ADD(csum, -b, base);
    ans += store[a];
    store[csum]++;
  }
  zeros += (long long)len * (len + 1) / 2;
  if (b == 0)
    cout << zeros << endl;
  else if (b == base - 1)
    cout << ans - zeros << endl;
  else
    cout << ans << endl;
  return 0;
}

```