#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
template <typename T>
void printV(vector<T>& v) {
  for (int i = 0; i < (int)(v.size()); i++) cout << v[i] << " ";
  cout << endl;
}
template <typename T>
void printVs(vector<vector<T> >& v) {
  for (int i = 0; i < (int)(v.size()); i++) printV(v[i]);
}
template <typename T>
void printVs(T** v, int H, int W) {
  for (int i = 0; i < (int)(H); i++) {
    for (int j = 0; j < (int)(W); j++) cout << v[i][j] << " ";
    cout << endl;
  }
}
void printV(vector<pair<int, int> >& v) {
  for (int i = 0; i < (int)(v.size()); i++)
    cout << "(" << v[i].first << "," << v[i].second << ") ";
}
void err() { cout << "wtf " << endl; }
vector<vector<int> > nxt;
vector<bool> seen;
long long bsearch(int idx, vector<int>& v) {
  int low = 0;
  int high = v.size();
  int ans = high;
  while (low < high) {
    int mid = (low + high) / 2;
    if (v[mid] > idx) {
      ans = mid;
      high = mid;
    } else
      low = mid + 1;
  }
  return v.size() - ans;
}
int main() {
  map<int, vector<int> > M;
  int k, b, n;
  cin >> k >> b >> n;
  vector<int> dg(n);
  k--;
  for (int i = 0; i < (int)(n); i++) cin >> dg[i];
  long long sum = 0;
  for (int i = 0; i < (int)(n); i++) {
    sum += dg[i];
    M[sum % k].push_back(i);
  }
  long long ttl = 0;
  sum = 0;
  for (int i = 0; i < (int)(n); i++) {
    sum += dg[i];
    int rem = (sum + b) % k;
    if (M.find(rem) == M.end()) continue;
    vector<int>& v = M[rem];
    ttl += bsearch(i, v);
  }
  ttl += bsearch(-1, M[b % k]);
  long long zero = 0;
  int st = 0;
  int end = st;
  for (int i = 0; i < (int)(n + 1); i++) {
    if (i == n || dg[i]) {
      long long len = end - st;
      st++;
      end = st;
      zero += ((len) * (len + 1)) / 2;
    } else
      end++;
  }
  if (b == 0)
    cout << zero << endl;
  else if (b == k)
    cout << ttl - zero << endl;
  else
    cout << ttl << endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long mod = 1000000007;
template <typename T>
void printV(vector<T>& v) {
  for (int i = 0; i < (int)(v.size()); i++) cout << v[i] << " ";
  cout << endl;
}
template <typename T>
void printVs(vector<vector<T> >& v) {
  for (int i = 0; i < (int)(v.size()); i++) printV(v[i]);
}
template <typename T>
void printVs(T** v, int H, int W) {
  for (int i = 0; i < (int)(H); i++) {
    for (int j = 0; j < (int)(W); j++) cout << v[i][j] << " ";
    cout << endl;
  }
}
void printV(vector<pair<int, int> >& v) {
  for (int i = 0; i < (int)(v.size()); i++)
    cout << "(" << v[i].first << "," << v[i].second << ") ";
}
void err() { cout << "wtf " << endl; }
vector<vector<int> > nxt;
vector<bool> seen;
long long bsearch(int idx, vector<int>& v) {
  int low = 0;
  int high = v.size();
  int ans = high;
  while (low < high) {
    int mid = (low + high) / 2;
    if (v[mid] > idx) {
      ans = mid;
      high = mid;
    } else
      low = mid + 1;
  }
  return v.size() - ans;
}
int main() {
  map<int, vector<int> > M;
  int k, b, n;
  cin >> k >> b >> n;
  vector<int> dg(n);
  k--;
  for (int i = 0; i < (int)(n); i++) cin >> dg[i];
  long long sum = 0;
  for (int i = 0; i < (int)(n); i++) {
    sum += dg[i];
    M[sum % k].push_back(i);
  }
  long long ttl = 0;
  sum = 0;
  for (int i = 0; i < (int)(n); i++) {
    sum += dg[i];
    int rem = (sum + b) % k;
    if (M.find(rem) == M.end()) continue;
    vector<int>& v = M[rem];
    ttl += bsearch(i, v);
  }
  ttl += bsearch(-1, M[b % k]);
  long long zero = 0;
  int st = 0;
  int end = st;
  for (int i = 0; i < (int)(n + 1); i++) {
    if (i == n || dg[i]) {
      long long len = end - st;
      st++;
      end = st;
      zero += ((len) * (len + 1)) / 2;
    } else
      end++;
  }
  if (b == 0)
    cout << zero << endl;
  else if (b == k)
    cout << ttl - zero << endl;
  else
    cout << ttl << endl;
}

```