#include <bits/stdc++.h>
using namespace std;
template <class F, class T>
T convert(F a, int p = -1) {
  stringstream ss;
  if (p >= 0) ss << fixed << setprecision(p);
  ss << a;
  T r;
  ss >> r;
  return r;
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T cube(T x) {
  return x * x * x;
}
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return s == 0 ? 0 : cntbit(s >> 1) + (s & 1);
}
template <class T>
int disp(T s) {
  for (int i = 0; i < (((int)(s).size())); ++i) cout << s[i] << " ";
  cout << '\n';
}
template <class T>
int disp(T s, int n) {
  for (int i = (0); i <= (n); ++i) cout << s[i] << " ";
  cout << '\n';
}
const long double PI = acos(-1.0);
const long double eps = 1e-9;
const int dr[] = {-1, 0, +1, 0};
const int dc[] = {0, +1, 0, -1};
const int inf = (int)1e9 + 5;
const long long linf = (long long)1e16 + 5;
const long long mod = (long long)1e9 + 7;
const int MX = 1e6 + 7;
long long ans;
int n, k, b, t;
int a[MX];
map<int, long long> M;
int sol() {
  cin >> k >> b >> n;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  if (b == k - 1) {
    for (int i = (1); i <= (n); ++i)
      if (a[i] == 0) {
        for (int j = (i + 1); j <= (n + 1); ++j)
          if (a[j] != 0 || j == n + 1) {
            ans -= (long long)(j - i) * (long long)(j + 1 - i) / 2;
            i = j - 1;
            break;
          }
      }
  }
  if (b == 0) {
    for (int i = (1); i <= (n); ++i)
      if (a[i] == 0) {
        for (int j = (i + 1); j <= (n + 1); ++j)
          if (a[j] != 0 || j == n + 1) {
            ans += (long long)(j - i) * (long long)(j + 1 - i) / 2;
            i = j - 1;
            break;
          }
      }
    cout << ans << '\n';
    return 0;
  }
  for (int i = (1); i <= (n); ++i) a[i] = (a[i] + a[i - 1]) % (k - 1);
  for (int i = (0); i <= (n); ++i) M[a[i]]++;
  for (int i = (0); i <= (n); ++i) {
    M[a[i]]--;
    t = (a[i] + b) % (k - 1);
    ans += M[t];
  }
  cout << ans << '\n';
}
int sltest = 1;
int main() {
  int xtest = 1;
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  while (xtest--) sol();
  return 0;
}

### Prompt

Generate a CPP solution 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>
using namespace std;
template <class F, class T>
T convert(F a, int p = -1) {
  stringstream ss;
  if (p >= 0) ss << fixed << setprecision(p);
  ss << a;
  T r;
  ss >> r;
  return r;
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T cube(T x) {
  return x * x * x;
}
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return s == 0 ? 0 : cntbit(s >> 1) + (s & 1);
}
template <class T>
int disp(T s) {
  for (int i = 0; i < (((int)(s).size())); ++i) cout << s[i] << " ";
  cout << '\n';
}
template <class T>
int disp(T s, int n) {
  for (int i = (0); i <= (n); ++i) cout << s[i] << " ";
  cout << '\n';
}
const long double PI = acos(-1.0);
const long double eps = 1e-9;
const int dr[] = {-1, 0, +1, 0};
const int dc[] = {0, +1, 0, -1};
const int inf = (int)1e9 + 5;
const long long linf = (long long)1e16 + 5;
const long long mod = (long long)1e9 + 7;
const int MX = 1e6 + 7;
long long ans;
int n, k, b, t;
int a[MX];
map<int, long long> M;
int sol() {
  cin >> k >> b >> n;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  if (b == k - 1) {
    for (int i = (1); i <= (n); ++i)
      if (a[i] == 0) {
        for (int j = (i + 1); j <= (n + 1); ++j)
          if (a[j] != 0 || j == n + 1) {
            ans -= (long long)(j - i) * (long long)(j + 1 - i) / 2;
            i = j - 1;
            break;
          }
      }
  }
  if (b == 0) {
    for (int i = (1); i <= (n); ++i)
      if (a[i] == 0) {
        for (int j = (i + 1); j <= (n + 1); ++j)
          if (a[j] != 0 || j == n + 1) {
            ans += (long long)(j - i) * (long long)(j + 1 - i) / 2;
            i = j - 1;
            break;
          }
      }
    cout << ans << '\n';
    return 0;
  }
  for (int i = (1); i <= (n); ++i) a[i] = (a[i] + a[i - 1]) % (k - 1);
  for (int i = (0); i <= (n); ++i) M[a[i]]++;
  for (int i = (0); i <= (n); ++i) {
    M[a[i]]--;
    t = (a[i] + b) % (k - 1);
    ans += M[t];
  }
  cout << ans << '\n';
}
int sltest = 1;
int main() {
  int xtest = 1;
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  while (xtest--) sol();
  return 0;
}

```