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".
#include <bits/stdc++.h>
using namespace std;
int k, b, n;
const int limite = 100000;
int a[limite];
long long int cuantos = 0;
void cuenta(int i1, int i2) {
  if (i1 == i2) {
    cuantos += a[i1] == b;
    return;
  }
  int med = (i1 + i2) / 2;
  cuenta(i1, med);
  cuenta(med + 1, i2);
  map<int, int> m;
  int num0 = 0;
  {
    long long int suma = 0;
    for (int i = med; i >= i1; i--) {
      suma += a[i];
      if (suma == 0)
        num0++;
      else
        m[suma % (k - 1)]++;
    }
  }
  {
    long long int suma = 0;
    for (int i = med + 1; i <= i2; i++) {
      suma += a[i];
      if (suma == 0) {
        if (b == 0)
          cuantos += num0;
        else
          cuantos += m[b % (k - 1)];
      } else if (b != 0) {
        cuantos += m[(b - suma % (k - 1) + k - 1) % (k - 1)];
        if (suma % (k - 1) == b % (k - 1)) cuantos += num0;
      }
    }
  }
}
int main() {
  cin >> k >> b >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  cuenta(0, n - 1);
  cout << cuantos << endl;
}
