#include <bits/stdc++.h>
const long maxn = 200000;
long i, j, k, n, m, b, t;
long x;
long long ans, kao;
long a[maxn], c[maxn];
long son[maxn * 20][2], gs[maxn * 20];
void find(long q, long l, long r, long x, long y) {
  long mid = (l + r) >> 1;
  long i = (x > mid);
  if (l == r) {
    kao = gs[q];
    kao *= y;
    ans += kao;
    return;
  }
  if (gs[q] == 0 || son[q][i] == 0) return;
  if (i)
    find(son[q][1], mid + 1, r, x, y);
  else
    find(son[q][0], l, mid, x, y);
  return;
}
void build(long q, long l, long r, long x, long y) {
  long mid = (l + r) >> 1;
  long i = (x > mid);
  gs[q] += y;
  if (l == r) return;
  if (son[q][i] == 0) son[q][i] = ++t;
  if (i)
    build(son[q][1], mid + 1, r, x, y);
  else
    build(son[q][0], l, mid, x, y);
  return;
}
int main() {
  scanf("%ld%ld%ld", &k, &b, &n);
  if (b >= k || b < 0) {
    puts("0");
    return 0;
  }
  if (b == 0) {
    for (i = 1; i <= n; ++i) {
      scanf("%ld", &x);
      if (x == 0) {
        ++j;
      } else {
        kao = j;
        kao *= j - 1;
        kao /= 2;
        kao += j;
        ans += kao;
        j = 0;
      }
    }
    if (j) {
      kao = j;
      kao *= j - 1;
      kao /= 2;
      kao += j;
      ans += kao;
    }
    printf("%I64d\n", ans);
    return 0;
  }
  --k;
  b %= k;
  t = 1;
  for (i = 1; i <= n; ++i) {
    scanf("%ld", &a[i]);
    if (a[i] == 0) {
      --i;
      --n;
      ++c[i];
    }
  }
  build(1, 0, k, 0, c[0] + 1);
  for (i = 1; i <= n; ++i) {
    a[i] += a[i - 1];
    a[i] %= k;
    find(1, 0, k, (a[i] - b + k) % k, c[i] + 1);
    build(1, 0, k, a[i], c[i] + 1);
  }
  printf("%I64d\n", ans);
  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>
const long maxn = 200000;
long i, j, k, n, m, b, t;
long x;
long long ans, kao;
long a[maxn], c[maxn];
long son[maxn * 20][2], gs[maxn * 20];
void find(long q, long l, long r, long x, long y) {
  long mid = (l + r) >> 1;
  long i = (x > mid);
  if (l == r) {
    kao = gs[q];
    kao *= y;
    ans += kao;
    return;
  }
  if (gs[q] == 0 || son[q][i] == 0) return;
  if (i)
    find(son[q][1], mid + 1, r, x, y);
  else
    find(son[q][0], l, mid, x, y);
  return;
}
void build(long q, long l, long r, long x, long y) {
  long mid = (l + r) >> 1;
  long i = (x > mid);
  gs[q] += y;
  if (l == r) return;
  if (son[q][i] == 0) son[q][i] = ++t;
  if (i)
    build(son[q][1], mid + 1, r, x, y);
  else
    build(son[q][0], l, mid, x, y);
  return;
}
int main() {
  scanf("%ld%ld%ld", &k, &b, &n);
  if (b >= k || b < 0) {
    puts("0");
    return 0;
  }
  if (b == 0) {
    for (i = 1; i <= n; ++i) {
      scanf("%ld", &x);
      if (x == 0) {
        ++j;
      } else {
        kao = j;
        kao *= j - 1;
        kao /= 2;
        kao += j;
        ans += kao;
        j = 0;
      }
    }
    if (j) {
      kao = j;
      kao *= j - 1;
      kao /= 2;
      kao += j;
      ans += kao;
    }
    printf("%I64d\n", ans);
    return 0;
  }
  --k;
  b %= k;
  t = 1;
  for (i = 1; i <= n; ++i) {
    scanf("%ld", &a[i]);
    if (a[i] == 0) {
      --i;
      --n;
      ++c[i];
    }
  }
  build(1, 0, k, 0, c[0] + 1);
  for (i = 1; i <= n; ++i) {
    a[i] += a[i - 1];
    a[i] %= k;
    find(1, 0, k, (a[i] - b + k) % k, c[i] + 1);
    build(1, 0, k, a[i], c[i] + 1);
  }
  printf("%I64d\n", ans);
  return 0;
}

```