#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <class T>
inline T max1(T a, T b) {
  return a > b ? a : b;
}
template <class T>
inline T min1(T a, T b) {
  return a < b ? a : b;
}
template <class T>
inline T gcd1(T a, T b) {
  if (a < b) swap(a, b);
  if (a % b == 0) return b;
  return gcd1(b, a % b);
}
struct node {
  int no;
  long long val;
  inline bool operator<(const node &a) const {
    if (val == a.val) return no < a.no;
    return val < a.val;
  }
};
long long jin, ans, n;
node dp[210000];
long long orig[100100];
int main() {
  scanf("%I64d%I64d%I64d", &jin, &ans, &n);
  long long temp = 0;
  long long ans1 = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &orig[i]);
    long long a = orig[i];
    if (a == 0)
      temp++;
    else {
      ans1 += temp * (temp + 1) / 2;
      temp = 0;
    }
  }
  ans1 += temp * (temp + 1) / 2;
  if (ans == 0) {
    cout << ans1 << endl;
    return 0;
  } else if ((jin == 2 && ans == 1)) {
    cout << n * (n + 1) / 2 - ans1 << endl;
    return 0;
  }
  long long ans0 = ans1;
  ans %= jin - 1;
  temp = 0;
  dp[0].val = 0;
  dp[0].no = 0;
  ans1 = 0;
  for (int i = 0; i < n; i++) {
    long long a = orig[i];
    temp += a;
    temp %= jin - 1;
    dp[i + 1].no = i + 1;
    dp[i + 1].val = temp;
  }
  sort(dp, dp + n + 1);
  for (int i = 0; i < n + 1; i++) {
    dp[i + n + 1] = dp[i];
    dp[i + n + 1].val += jin - 1;
  }
  int link1 = 0, link2 = 0, linkno = 0;
  while (dp[link2].val < (dp[link1].val + ans)) link2++;
  int end = link2 + n + 1;
  while (dp[link1].val + ans < dp[link2].val) link1++;
  linkno = link1 - 1;
  for (; link2 != end; link2++) {
    while (dp[link1].no < dp[link2].no && dp[link1].val + ans == dp[link2].val)
      link1++;
    ans1 += link1 - linkno - 1;
    if (dp[link2 + 1].val != dp[link2].val) {
      while (dp[link1].val + ans < dp[link2 + 1].val) link1++;
      linkno = link1 - 1;
    }
  }
  if (ans == 0)
    cout << ans1 - ans0 << endl;
  else
    cout << ans1 << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <class T>
inline T max1(T a, T b) {
  return a > b ? a : b;
}
template <class T>
inline T min1(T a, T b) {
  return a < b ? a : b;
}
template <class T>
inline T gcd1(T a, T b) {
  if (a < b) swap(a, b);
  if (a % b == 0) return b;
  return gcd1(b, a % b);
}
struct node {
  int no;
  long long val;
  inline bool operator<(const node &a) const {
    if (val == a.val) return no < a.no;
    return val < a.val;
  }
};
long long jin, ans, n;
node dp[210000];
long long orig[100100];
int main() {
  scanf("%I64d%I64d%I64d", &jin, &ans, &n);
  long long temp = 0;
  long long ans1 = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &orig[i]);
    long long a = orig[i];
    if (a == 0)
      temp++;
    else {
      ans1 += temp * (temp + 1) / 2;
      temp = 0;
    }
  }
  ans1 += temp * (temp + 1) / 2;
  if (ans == 0) {
    cout << ans1 << endl;
    return 0;
  } else if ((jin == 2 && ans == 1)) {
    cout << n * (n + 1) / 2 - ans1 << endl;
    return 0;
  }
  long long ans0 = ans1;
  ans %= jin - 1;
  temp = 0;
  dp[0].val = 0;
  dp[0].no = 0;
  ans1 = 0;
  for (int i = 0; i < n; i++) {
    long long a = orig[i];
    temp += a;
    temp %= jin - 1;
    dp[i + 1].no = i + 1;
    dp[i + 1].val = temp;
  }
  sort(dp, dp + n + 1);
  for (int i = 0; i < n + 1; i++) {
    dp[i + n + 1] = dp[i];
    dp[i + n + 1].val += jin - 1;
  }
  int link1 = 0, link2 = 0, linkno = 0;
  while (dp[link2].val < (dp[link1].val + ans)) link2++;
  int end = link2 + n + 1;
  while (dp[link1].val + ans < dp[link2].val) link1++;
  linkno = link1 - 1;
  for (; link2 != end; link2++) {
    while (dp[link1].no < dp[link2].no && dp[link1].val + ans == dp[link2].val)
      link1++;
    ans1 += link1 - linkno - 1;
    if (dp[link2 + 1].val != dp[link2].val) {
      while (dp[link1].val + ans < dp[link2 + 1].val) link1++;
      linkno = link1 - 1;
    }
  }
  if (ans == 0)
    cout << ans1 - ans0 << endl;
  else
    cout << ans1 << endl;
  return 0;
}

```