#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
inline unsigned int myrand(int n) {
  unsigned int u = rand() % n;
  int c = rand() % 10;
  u ^= u << c;
  u ^= (u << 7) & 263692864;
  u ^= (u << 15) & 402273075;
  u %= n;
  return u;
}
int sum[100011];
int a[100011];
map<int, int> cnt;
int main() {
  int k, b, n;
  cin >> k >> b >> n;
  k--;
  long long ans = 0;
  if (b == 0) {
    for (int i = 1; i <= n; i++) {
      gn(a[i]);
    }
    for (int i = 1, j = 1; i <= n; i = j) {
      if (a[j]) {
        j++;
        continue;
      }
      while (j <= n and a[j] == a[i]) j++;
      ans += (long long)(j - i) * (j - i + 1) / 2;
    }
    cout << ans << endl;
    return 0;
  }
  if (b == k) b = 0;
  cnt[0]++;
  for (int i = 1; i <= n; i++) {
    gn(a[i]);
    sum[i] = (sum[i - 1] + a[i]) % k;
    ans += cnt[(sum[i] - b + k) % k];
    cnt[sum[i]]++;
  }
  if (b == 0)
    for (int i = 1, j = 1; i <= n; i = j) {
      if (a[j]) {
        j++;
        continue;
      }
      while (j <= n and a[j] == a[i]) j++;
      ans -= (long long)(j - i) * (j - i + 1) / 2;
    }
  cout << ans << endl;
  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 <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
inline unsigned int myrand(int n) {
  unsigned int u = rand() % n;
  int c = rand() % 10;
  u ^= u << c;
  u ^= (u << 7) & 263692864;
  u ^= (u << 15) & 402273075;
  u %= n;
  return u;
}
int sum[100011];
int a[100011];
map<int, int> cnt;
int main() {
  int k, b, n;
  cin >> k >> b >> n;
  k--;
  long long ans = 0;
  if (b == 0) {
    for (int i = 1; i <= n; i++) {
      gn(a[i]);
    }
    for (int i = 1, j = 1; i <= n; i = j) {
      if (a[j]) {
        j++;
        continue;
      }
      while (j <= n and a[j] == a[i]) j++;
      ans += (long long)(j - i) * (j - i + 1) / 2;
    }
    cout << ans << endl;
    return 0;
  }
  if (b == k) b = 0;
  cnt[0]++;
  for (int i = 1; i <= n; i++) {
    gn(a[i]);
    sum[i] = (sum[i - 1] + a[i]) % k;
    ans += cnt[(sum[i] - b + k) % k];
    cnt[sum[i]]++;
  }
  if (b == 0)
    for (int i = 1, j = 1; i <= n; i = j) {
      if (a[j]) {
        j++;
        continue;
      }
      while (j <= n and a[j] == a[i]) j++;
      ans -= (long long)(j - i) * (j - i + 1) / 2;
    }
  cout << ans << endl;
  return 0;
}

```