216_E. Martian Luck_13582
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".
from bisect import *
from itertools import islice

k, b, n = [int(x) for x in raw_input().split()]
nums = [int(x) for x in raw_input().split()]
km1 = k - 1
sums = [0]
for x in nums:
    sums.append((x + sums[-1]) % km1)

sums_with_idx = [(x, i) for i, x in enumerate(sums)]
sums_with_idx.sort()

ans = 0
for i, x in enumerate(islice(sums, 0, n)):
    t = (b + x) % km1
    ans += bisect_right(sums_with_idx, (t, n)) - bisect_left(sums_with_idx, (t, i+1))

if b != km1 and b != 0:
    print ans
else:
    zeros_idx_left = [i for i, x in enumerate(nums) if x == 0 and (i == 0 or nums[i-1] != 0)]
    zeros_idx_right = [i for i, x in enumerate(nums) if x == 0 and (i == n-1 or nums[i+1] != 0)]
    cont_zeros = [y - x + 1 for x, y in zip(zeros_idx_left, zeros_idx_right)]
    zeros = 0 if len(cont_zeros) == 0 else reduce(lambda x, y : x + (1 + y) * y / 2, cont_zeros[1:], cont_zeros[0] * (cont_zeros[0] + 1) / 2)
    if b == 0:
        print zeros
    else:
        print ans - zeros
