#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void input(T &x) {
  register char c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}
inline long long bigmod(long long p, long long e, long long M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
const int N = 1e5 + 5;
long long k, b, n;
long long a[N];
long long d(long long n) {
  while (n >= k) {
    long long nn = 0;
    while (n) nn += n % k, n /= k;
    n = nn;
  }
  return n;
}
void f() {
  while (1) {
    cin >> k >> b;
    for (int i = 0; i <= 1000; i++)
      if (d(i) == b) cout << i << " ";
    cout << endl;
  }
}
int MOD;
int main() {
  cin >> k >> b >> n;
  MOD = k - 1;
  for (int i = 1; i <= n; i++) cin >> a[i];
  long long ans2 = 0, cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 0)
      cnt++;
    else
      ans2 += cnt * (cnt + 1) / 2, cnt = 0;
  }
  ans2 += cnt * (cnt + 1) / 2;
  if (b == 0) {
    cout << ans2 << endl;
    return 0;
  }
  long long ans = 0;
  long long add = 0;
  map<long long, long long> Map;
  for (int i = n; i >= 1; i--) {
    add += a[i];
    add %= MOD;
    long long now = ((a[i] - add) % MOD + MOD) % MOD;
    Map[now]++;
    int v = (b - add + MOD) % MOD;
    if (Map.count(v)) ans += Map[v];
  }
  cout << ans - ((b == k - 1) ? ans2 : 0) << endl;
  return 0;
}

### Prompt

Your task is to create 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 T>
inline void input(T &x) {
  register char c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}
inline long long bigmod(long long p, long long e, long long M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
const int N = 1e5 + 5;
long long k, b, n;
long long a[N];
long long d(long long n) {
  while (n >= k) {
    long long nn = 0;
    while (n) nn += n % k, n /= k;
    n = nn;
  }
  return n;
}
void f() {
  while (1) {
    cin >> k >> b;
    for (int i = 0; i <= 1000; i++)
      if (d(i) == b) cout << i << " ";
    cout << endl;
  }
}
int MOD;
int main() {
  cin >> k >> b >> n;
  MOD = k - 1;
  for (int i = 1; i <= n; i++) cin >> a[i];
  long long ans2 = 0, cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 0)
      cnt++;
    else
      ans2 += cnt * (cnt + 1) / 2, cnt = 0;
  }
  ans2 += cnt * (cnt + 1) / 2;
  if (b == 0) {
    cout << ans2 << endl;
    return 0;
  }
  long long ans = 0;
  long long add = 0;
  map<long long, long long> Map;
  for (int i = n; i >= 1; i--) {
    add += a[i];
    add %= MOD;
    long long now = ((a[i] - add) % MOD + MOD) % MOD;
    Map[now]++;
    int v = (b - add + MOD) % MOD;
    if (Map.count(v)) ans += Map[v];
  }
  cout << ans - ((b == k - 1) ? ans2 : 0) << endl;
  return 0;
}

```