#include <bits/stdc++.h>
inline unsigned int getuint() {
  char w = getchar();
  while (w < '0' || '9' < w) w = getchar();
  unsigned int ans = 0;
  for (; '0' <= w && w <= '9'; w = getchar()) ans = ans * 10 + w - '0';
  return ans;
}
int k, m, n;
int a[100005];
long long cal0() {
  long long ans = 0;
  for (int l = 1, r; l <= n; l = r + 1)
    if (!a[r = l]) {
      for (; r < n && !a[r + 1]; r++)
        ;
      long long len = r - l + 1;
      ans += (1 + len) * len >> 1;
    }
  return ans;
}
std::map<int, int> cnt;
int main() {
  k = getuint(), m = getuint(), n = getuint();
  int mod = k - 1;
  for (int i = 1; i <= n; i++) a[i] = getuint();
  long long ans = cal0();
  if (!m) return printf("%I64d\n", ans), 0;
  if (m == mod)
    m = 0, ans = -ans;
  else
    ans = 0;
  cnt[0] = 1;
  int x = 0;
  for (int i = 1; i <= n; i++) {
    x = (x + a[i]) % mod;
    int t = (x - m + mod) % mod;
    if (cnt.find(t) != cnt.end()) ans += cnt[t];
    if (cnt.find(x) == cnt.end()) cnt[x] = 0;
    cnt[x]++;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please create a solution in CPP to 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>
inline unsigned int getuint() {
  char w = getchar();
  while (w < '0' || '9' < w) w = getchar();
  unsigned int ans = 0;
  for (; '0' <= w && w <= '9'; w = getchar()) ans = ans * 10 + w - '0';
  return ans;
}
int k, m, n;
int a[100005];
long long cal0() {
  long long ans = 0;
  for (int l = 1, r; l <= n; l = r + 1)
    if (!a[r = l]) {
      for (; r < n && !a[r + 1]; r++)
        ;
      long long len = r - l + 1;
      ans += (1 + len) * len >> 1;
    }
  return ans;
}
std::map<int, int> cnt;
int main() {
  k = getuint(), m = getuint(), n = getuint();
  int mod = k - 1;
  for (int i = 1; i <= n; i++) a[i] = getuint();
  long long ans = cal0();
  if (!m) return printf("%I64d\n", ans), 0;
  if (m == mod)
    m = 0, ans = -ans;
  else
    ans = 0;
  cnt[0] = 1;
  int x = 0;
  for (int i = 1; i <= n; i++) {
    x = (x + a[i]) % mod;
    int t = (x - m + mod) % mod;
    if (cnt.find(t) != cnt.end()) ans += cnt[t];
    if (cnt.find(x) == cnt.end()) cnt[x] = 0;
    cnt[x]++;
  }
  printf("%I64d\n", ans);
  return 0;
}

```