#include <bits/stdc++.h>
using namespace std;
long long MD;
long long k, b, n, a[102030];
long long iPow(long long x, long long y) {
  long long ret = 1;
  for (long long j = 1; j <= y; j <<= 1) {
    if (j & y) ret = (long long)ret * x % MD;
    x = (long long)x * x % MD;
  }
  return ret;
}
long long phi(long long p) {
  long long ret = p;
  for (long long i = 2; i <= (int)sqrt(p + 0.000001); ++i)
    while (p % i == 0) {
      ret = ret / i * (i - 1);
      p /= i;
    }
  return ret;
}
map<long long, long long> MAP;
int main() {
  cin >> k >> b >> n;
  long long ans = 0;
  long long len = 0;
  for (long long i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    if (a[i] != 0) {
      ans += len * (len + 1) / 2;
      len = 0;
    } else
      ++len;
  }
  ans += len * (len + 1) / 2;
  if (b == 0) {
    cout << ans << endl;
    return 0;
  }
  long long tA = ans;
  ans = 0;
  if (b < k - 1) tA = 0;
  MD = k - 1;
  MAP.clear();
  long long inv = iPow(k, phi(MD) - 1);
  pair<int, int> chg = make_pair(0, 0);
  ans = 0;
  long long tB = b;
  b = (b + MD - 1) % MD;
  for (long long i = 1; i <= n + 1; ++i) {
    if (i == 3) {
      long long z = i;
    }
    if (MAP.count((b + 1) % (k - 1))) {
      ans += MAP[(b + 1) % (k - 1)];
    }
    long long t = ((long long)b * iPow(k, chg.first) % MD + chg.second) % MD;
    b = (b + MD - a[i]) % MD;
    b = (long long)b * inv % MD;
    chg.first++;
    chg.second = ((long long)chg.second * k + a[i]) % MD;
    long long tPos =
        ((long long)a[i] + MD - chg.second) % MD * iPow(inv, chg.first) % MD;
    ++MAP[tPos];
  }
  cout << ans - tA << endl;
  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;
long long MD;
long long k, b, n, a[102030];
long long iPow(long long x, long long y) {
  long long ret = 1;
  for (long long j = 1; j <= y; j <<= 1) {
    if (j & y) ret = (long long)ret * x % MD;
    x = (long long)x * x % MD;
  }
  return ret;
}
long long phi(long long p) {
  long long ret = p;
  for (long long i = 2; i <= (int)sqrt(p + 0.000001); ++i)
    while (p % i == 0) {
      ret = ret / i * (i - 1);
      p /= i;
    }
  return ret;
}
map<long long, long long> MAP;
int main() {
  cin >> k >> b >> n;
  long long ans = 0;
  long long len = 0;
  for (long long i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    if (a[i] != 0) {
      ans += len * (len + 1) / 2;
      len = 0;
    } else
      ++len;
  }
  ans += len * (len + 1) / 2;
  if (b == 0) {
    cout << ans << endl;
    return 0;
  }
  long long tA = ans;
  ans = 0;
  if (b < k - 1) tA = 0;
  MD = k - 1;
  MAP.clear();
  long long inv = iPow(k, phi(MD) - 1);
  pair<int, int> chg = make_pair(0, 0);
  ans = 0;
  long long tB = b;
  b = (b + MD - 1) % MD;
  for (long long i = 1; i <= n + 1; ++i) {
    if (i == 3) {
      long long z = i;
    }
    if (MAP.count((b + 1) % (k - 1))) {
      ans += MAP[(b + 1) % (k - 1)];
    }
    long long t = ((long long)b * iPow(k, chg.first) % MD + chg.second) % MD;
    b = (b + MD - a[i]) % MD;
    b = (long long)b * inv % MD;
    chg.first++;
    chg.second = ((long long)chg.second * k + a[i]) % MD;
    long long tPos =
        ((long long)a[i] + MD - chg.second) % MD * iPow(inv, chg.first) % MD;
    ++MAP[tPos];
  }
  cout << ans - tA << endl;
  return 0;
}

```