#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const double EPS = 1e-10;
int in() {
  int r = 0, c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') return -in();
  for (; c > 32; r = (r << 1) + (r << 3) + c - '0', c = getchar())
    ;
  return r;
}
long long A[101000];
long long acum[101000];
void solve() {
  int K = in();
  int B = in();
  int N = in();
  int M = K - 1;
  int i;
  long long voy = 0;
  acum[0] = 0;
  for (i = 1; i <= N; i++) {
    A[i] = in();
    if (A[i] == 0) {
      acum[i] = acum[i - 1] + 1;
    } else {
      acum[i] = 0;
    }
  }
  acum[N + 1] = 0;
  long long ceroes = 0;
  for (i = 1; i <= N + 1; i++) {
    if (acum[i] == 0) ceroes += ((acum[i - 1]) * (acum[i - 1] + 1) / 2);
  }
  if (B == 0) {
    cout << ceroes << endl;
    return;
  }
  long long ans = 0;
  map<int, long long> pref;
  pref[0] = 1;
  voy = 0;
  for (i = 1; i <= N; i++) {
    voy += A[i];
    ans += pref[(((voy % M - B) % M) + M) % M];
    pref[voy % M]++;
  }
  if (B == M) ans -= ceroes;
  cout << ans << endl;
}
int main() {
  solve();
  return 0;
}

### 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;
const double PI = acos(-1.0);
const double EPS = 1e-10;
int in() {
  int r = 0, c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') return -in();
  for (; c > 32; r = (r << 1) + (r << 3) + c - '0', c = getchar())
    ;
  return r;
}
long long A[101000];
long long acum[101000];
void solve() {
  int K = in();
  int B = in();
  int N = in();
  int M = K - 1;
  int i;
  long long voy = 0;
  acum[0] = 0;
  for (i = 1; i <= N; i++) {
    A[i] = in();
    if (A[i] == 0) {
      acum[i] = acum[i - 1] + 1;
    } else {
      acum[i] = 0;
    }
  }
  acum[N + 1] = 0;
  long long ceroes = 0;
  for (i = 1; i <= N + 1; i++) {
    if (acum[i] == 0) ceroes += ((acum[i - 1]) * (acum[i - 1] + 1) / 2);
  }
  if (B == 0) {
    cout << ceroes << endl;
    return;
  }
  long long ans = 0;
  map<int, long long> pref;
  pref[0] = 1;
  voy = 0;
  for (i = 1; i <= N; i++) {
    voy += A[i];
    ans += pref[(((voy % M - B) % M) + M) % M];
    pref[voy % M]++;
  }
  if (B == M) ans -= ceroes;
  cout << ans << endl;
}
int main() {
  solve();
  return 0;
}

```