#include <bits/stdc++.h>
using namespace std;
long long spf[300001];
long long fac[300001];
void sieve() {
  spf[1] = 1;
  for (long long i = 2; i < 300001; i++) spf[i] = i;
  for (long long i = 4; i < 300001; i += 2) spf[i] = 2;
  for (long long i = 3; i * i < 300001; i++) {
    if (spf[i] == i) {
      for (long long j = i * i; j < 300001; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
map<long long, long long> getfactor(long long a) {
  map<long long, long long> m;
  while (a > 1) {
    m[spf[a]]++;
    a /= spf[a];
  }
  return m;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long inverse(long long a, long long p) { return power(a, p - 2, p); }
long long ncr(long long n, long long r, long long p) {
  if (r == 0) return 1;
  return (fac[n] * inverse(fac[r], p) % p * inverse(fac[n - r], p) % p) % p;
}
void solve() {
  long long k, b, n;
  cin >> k >> b >> n;
  long long a[n + 1];
  for (long long i = 1; i <= n; i++) cin >> a[i];
  map<long long, long long> m;
  m[0]++;
  long long tot = 0;
  long long ans = 0;
  long long ans1 = 0;
  long long sum1 = 0;
  for (long long i = 1; i <= n; i++) {
    if (a[i] == 0)
      sum1++;
    else {
      ans1 += (sum1 * sum1 + sum1) / 2;
      sum1 = 0;
    }
  }
  ans1 += (sum1 * sum1 + sum1) / 2;
  for (long long i = 1; i <= n; i++) {
    tot += a[i];
    long long sum = tot;
    while (sum >= k) {
      long long temp = sum;
      long long val = 0;
      while (temp > 0) {
        val += temp % k;
        temp /= k;
      }
      sum = val;
    }
    if (sum >= b) ans += m[sum - b];
    ans += m[sum + k - 1 - b];
    m[sum]++;
  }
  if (b == 0)
    cout << ans1 << '\n';
  else if (b == k - 1)
    cout << ans - ans1 << '\n';
  else
    cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long test = 1;
  while (test--) {
    solve();
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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 spf[300001];
long long fac[300001];
void sieve() {
  spf[1] = 1;
  for (long long i = 2; i < 300001; i++) spf[i] = i;
  for (long long i = 4; i < 300001; i += 2) spf[i] = 2;
  for (long long i = 3; i * i < 300001; i++) {
    if (spf[i] == i) {
      for (long long j = i * i; j < 300001; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
map<long long, long long> getfactor(long long a) {
  map<long long, long long> m;
  while (a > 1) {
    m[spf[a]]++;
    a /= spf[a];
  }
  return m;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long inverse(long long a, long long p) { return power(a, p - 2, p); }
long long ncr(long long n, long long r, long long p) {
  if (r == 0) return 1;
  return (fac[n] * inverse(fac[r], p) % p * inverse(fac[n - r], p) % p) % p;
}
void solve() {
  long long k, b, n;
  cin >> k >> b >> n;
  long long a[n + 1];
  for (long long i = 1; i <= n; i++) cin >> a[i];
  map<long long, long long> m;
  m[0]++;
  long long tot = 0;
  long long ans = 0;
  long long ans1 = 0;
  long long sum1 = 0;
  for (long long i = 1; i <= n; i++) {
    if (a[i] == 0)
      sum1++;
    else {
      ans1 += (sum1 * sum1 + sum1) / 2;
      sum1 = 0;
    }
  }
  ans1 += (sum1 * sum1 + sum1) / 2;
  for (long long i = 1; i <= n; i++) {
    tot += a[i];
    long long sum = tot;
    while (sum >= k) {
      long long temp = sum;
      long long val = 0;
      while (temp > 0) {
        val += temp % k;
        temp /= k;
      }
      sum = val;
    }
    if (sum >= b) ans += m[sum - b];
    ans += m[sum + k - 1 - b];
    m[sum]++;
  }
  if (b == 0)
    cout << ans1 << '\n';
  else if (b == k - 1)
    cout << ans - ans1 << '\n';
  else
    cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long test = 1;
  while (test--) {
    solve();
  }
}

```