#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-8;
inline int calc(int x, int base) {
  while (x >= base) {
    int y = 0;
    while (x > 0) {
      y += x % base;
      x /= base;
    }
    x = y;
  }
  return x;
}
const int maxn = 100005;
char s[1000000];
int cnt[maxn], a[maxn], bak[maxn], n, lucky, base;
int main() {
  scanf("%d%d%d", &base, &lucky, &n);
  int cur = 0, zero = 0;
  long long ans0 = 0;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    if (x == 0)
      ++zero;
    else
      zero = 0;
    ans0 += zero;
    cur += x;
    cur %= base - 1;
    a[i] = cur;
    bak[i] = cur;
  }
  bak[n] = 0;
  sort(bak, bak + n + 1);
  int m = unique(bak, bak + n + 1) - bak;
  memset(cnt, 0, sizeof(cnt));
  ++cnt[0];
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    int x = a[i] - lucky;
    if (x < 0) x += base - 1;
    int index = lower_bound(bak, bak + m, x) - bak;
    if (index < m && bak[index] == x) {
      ans += cnt[index];
    }
    index = lower_bound(bak, bak + m, a[i]) - bak;
    ++cnt[index];
  }
  if (lucky == 0)
    ans = ans0;
  else if (lucky == base - 1)
    ans -= ans0;
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const double pi = acos(-1.0);
const double eps = 1e-8;
inline int calc(int x, int base) {
  while (x >= base) {
    int y = 0;
    while (x > 0) {
      y += x % base;
      x /= base;
    }
    x = y;
  }
  return x;
}
const int maxn = 100005;
char s[1000000];
int cnt[maxn], a[maxn], bak[maxn], n, lucky, base;
int main() {
  scanf("%d%d%d", &base, &lucky, &n);
  int cur = 0, zero = 0;
  long long ans0 = 0;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    if (x == 0)
      ++zero;
    else
      zero = 0;
    ans0 += zero;
    cur += x;
    cur %= base - 1;
    a[i] = cur;
    bak[i] = cur;
  }
  bak[n] = 0;
  sort(bak, bak + n + 1);
  int m = unique(bak, bak + n + 1) - bak;
  memset(cnt, 0, sizeof(cnt));
  ++cnt[0];
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    int x = a[i] - lucky;
    if (x < 0) x += base - 1;
    int index = lower_bound(bak, bak + m, x) - bak;
    if (index < m && bak[index] == x) {
      ans += cnt[index];
    }
    index = lower_bound(bak, bak + m, a[i]) - bak;
    ++cnt[index];
  }
  if (lucky == 0)
    ans = ans0;
  else if (lucky == base - 1)
    ans -= ans0;
  cout << ans << endl;
  return 0;
}

```