#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
int n, k, b, ik, a[N], pw[N];
long long ans = 0;
long long ext(long long a, long long b, long long &x, long long &y) {
  if (b == 0) return x = 1, y = 0, a;
  long long t, g = ext(b, a % b, x, y);
  t = x, x = y, y = t - a / b * y;
  return g;
}
long long invmod(long long a, long long n) {
  long long x, y;
  if (ext(a, n, x, y) != 1) return -1;
  return (x % n + n) % n;
}
long long kpow(long long a, long long b, long long mod) {
  long long r = 1;
  while (b > 0) {
    if (b & 1) r = r * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return r;
}
void solve(int l, int r) {
  if (l == r) {
    ans += a[l] == b;
    return;
  }
  int m = (l + r) >> 1;
  solve(l, m), solve(m + 1, r);
  map<int, int> S;
  int x = 0;
  for (int i = m; i >= l; --i) {
    x = (1ll * x * k + a[i]) % (k - 1);
    if (S.find(x) == S.end()) S[x] = 0;
    ++S[x];
  }
  int y = 0;
  for (int i = (m + 1), I = (r + 1); i < I; ++i) {
    y = (1ll * y * k + a[i]) % (k - 1);
    x = 1ll * (b - y + k - 1) * pw[i - m] % (k - 1);
    if (S.find(x) != S.end()) ans += S[x];
  }
}
long long calc(long long c1, long long c2) {
  long long ret = 0;
  for (int i = 0, j; i < n; i = j) {
    j = i + 1;
    if (a[i] == 0) {
      while (j < n && !a[j]) ++j;
      int x = j - i;
      ret += c1 * x + (c2 * x * (x - 1) >> 1);
    }
  }
  return ret;
}
int main() {
  scanf("%d%d%d", &k, &b, &n);
  ik = invmod(k, k - 1);
  pw[0] = 1;
  for (int i = (1), I = (n + 1); i < I; ++i)
    pw[i] = 1ll * pw[i - 1] * ik % (k - 1);
  for (int i = (0), I = (n); i < I; ++i) scanf("%d", &a[i]);
  ans = 0;
  if (b == 0) {
    ans = calc(1, 1);
  } else {
    solve(0, n - 1);
    if (b == k - 1) ans -= calc(0, 1);
  }
  printf("%I64d", ans);
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 1e5 + 7;
int n, k, b, ik, a[N], pw[N];
long long ans = 0;
long long ext(long long a, long long b, long long &x, long long &y) {
  if (b == 0) return x = 1, y = 0, a;
  long long t, g = ext(b, a % b, x, y);
  t = x, x = y, y = t - a / b * y;
  return g;
}
long long invmod(long long a, long long n) {
  long long x, y;
  if (ext(a, n, x, y) != 1) return -1;
  return (x % n + n) % n;
}
long long kpow(long long a, long long b, long long mod) {
  long long r = 1;
  while (b > 0) {
    if (b & 1) r = r * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return r;
}
void solve(int l, int r) {
  if (l == r) {
    ans += a[l] == b;
    return;
  }
  int m = (l + r) >> 1;
  solve(l, m), solve(m + 1, r);
  map<int, int> S;
  int x = 0;
  for (int i = m; i >= l; --i) {
    x = (1ll * x * k + a[i]) % (k - 1);
    if (S.find(x) == S.end()) S[x] = 0;
    ++S[x];
  }
  int y = 0;
  for (int i = (m + 1), I = (r + 1); i < I; ++i) {
    y = (1ll * y * k + a[i]) % (k - 1);
    x = 1ll * (b - y + k - 1) * pw[i - m] % (k - 1);
    if (S.find(x) != S.end()) ans += S[x];
  }
}
long long calc(long long c1, long long c2) {
  long long ret = 0;
  for (int i = 0, j; i < n; i = j) {
    j = i + 1;
    if (a[i] == 0) {
      while (j < n && !a[j]) ++j;
      int x = j - i;
      ret += c1 * x + (c2 * x * (x - 1) >> 1);
    }
  }
  return ret;
}
int main() {
  scanf("%d%d%d", &k, &b, &n);
  ik = invmod(k, k - 1);
  pw[0] = 1;
  for (int i = (1), I = (n + 1); i < I; ++i)
    pw[i] = 1ll * pw[i - 1] * ik % (k - 1);
  for (int i = (0), I = (n); i < I; ++i) scanf("%d", &a[i]);
  ans = 0;
  if (b == 0) {
    ans = calc(1, 1);
  } else {
    solve(0, n - 1);
    if (b == k - 1) ans -= calc(0, 1);
  }
  printf("%I64d", ans);
  return 0;
}

```