#include <bits/stdc++.h>
using namespace std;
template <typename T>
using vc = vector<T>;
using pii = pair<long long, long long>;
void xmax(long long& a, long long b) { a = max(a, b); }
void xmin(long long& a, long long b) { a = min(a, b); }
void print(vc<long long>& a, string second) {
  cerr << "prinitng " << second << "\n";
  for (long long i = 0; i < a.size(); i++) {
    cerr << a[i] << " ";
  }
  cerr << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long k, b, n;
  map<long long, long long> anz;
  map<long long, long long> anz_null;
  anz_null[0] = 1;
  anz[0] = 1;
  cin >> k >> b >> n;
  k--;
  long long bb = b;
  b = b % k;
  long long sum = 0;
  long long cur = 0;
  vc<long long> aa(n);
  for (long long i = 0; i < n; i++) {
    cin >> aa[i];
  }
  long long cur_sum = 0;
  long long null_sum = 0;
  for (long long i = 0; i < n; i++) {
    long long a = aa[i];
    cur = (cur + a) % k;
    cur_sum += a;
    long long last = (cur + k - b) % k;
    long long add = anz[last];
    sum += add;
    long long add_null = anz_null[cur_sum];
    null_sum += add_null;
    anz[cur]++;
    anz_null[cur_sum]++;
  }
  if (bb == k) {
    cout << sum - null_sum << "\n";
  } else if (bb == 0) {
    cout << null_sum << "\n";
  } else
    cout << sum << "\n";
}

### 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;
template <typename T>
using vc = vector<T>;
using pii = pair<long long, long long>;
void xmax(long long& a, long long b) { a = max(a, b); }
void xmin(long long& a, long long b) { a = min(a, b); }
void print(vc<long long>& a, string second) {
  cerr << "prinitng " << second << "\n";
  for (long long i = 0; i < a.size(); i++) {
    cerr << a[i] << " ";
  }
  cerr << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long k, b, n;
  map<long long, long long> anz;
  map<long long, long long> anz_null;
  anz_null[0] = 1;
  anz[0] = 1;
  cin >> k >> b >> n;
  k--;
  long long bb = b;
  b = b % k;
  long long sum = 0;
  long long cur = 0;
  vc<long long> aa(n);
  for (long long i = 0; i < n; i++) {
    cin >> aa[i];
  }
  long long cur_sum = 0;
  long long null_sum = 0;
  for (long long i = 0; i < n; i++) {
    long long a = aa[i];
    cur = (cur + a) % k;
    cur_sum += a;
    long long last = (cur + k - b) % k;
    long long add = anz[last];
    sum += add;
    long long add_null = anz_null[cur_sum];
    null_sum += add_null;
    anz[cur]++;
    anz_null[cur_sum]++;
  }
  if (bb == k) {
    cout << sum - null_sum << "\n";
  } else if (bb == 0) {
    cout << null_sum << "\n";
  } else
    cout << sum << "\n";
}

```