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... a_n (1 ≤ i ≤ j ≤ n) is the string a_ia_i + 1... a_j. 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 a_i (0 ≤ a_i < 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.

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".

Idx	Name	Rating
A	Tiling with Hexagons	1200
B	Forming Teams	1700
C	Hiring Staff	1800
D	Spider's Web	1700
E	Martian Luck	2000

Codeforces Round 133 (Div. 2)