Problem:
Score : 500 points

Problem Statement

Takahashi has a string S of length N consisting of digits from 0 through 9.

He loves the prime number P. He wants to know how many non-empty (contiguous) substrings of S - there are N \times (N + 1) / 2 of them - are divisible by P when regarded as integers written in base ten.

Here substrings starting with a 0 also count, and substrings originated from different positions in S are distinguished, even if they are equal as strings or integers.

Compute this count to help Takahashi.

Constraints

1 \leq N \leq 2 \times 10^5

S consists of digits.

|S| = N

2 \leq P \leq 10000

P is a prime number.

Input

Input is given from Standard Input in the following format:

N P
S

Output

Print the number of non-empty (contiguous) substrings of S that are divisible by P when regarded as an integer written in base ten.

Sample Input 1

4 3
3543

Sample Output 1

6

Here S = 3543. There are ten non-empty (contiguous) substrings of S:

3: divisible by 3.

35: not divisible by 3.

354: divisible by 3.

3543: divisible by 3.

5: not divisible by 3.

54: divisible by 3.

543: divisible by 3.

4: not divisible by 3.

43: not divisible by 3.

3: divisible by 3.

Six of these are divisible by 3, so print 6.

Sample Input 2

4 2
2020

Sample Output 2

10

Here S = 2020. There are ten non-empty (contiguous) substrings of S, all of which are divisible by 2, so print 10.

Note that substrings beginning with a 0 also count.

Sample Input 3

20 11
33883322005544116655

Sample Output 3

68

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

n, p = gets.split.map(&:to_i)
s = gets.chop
l_s = s.length
l_p = p.to_s.length
ans = {}
current_top = 0
count = 0
loop do
  current_bottom = current_top + l_p - 1
  offset = 0
  loop do
    ts = s[(current_top + offset)..current_bottom].sub(/^0+/, '')
    if ans[ts.to_sym]
      count += 1
      offset += current_bottom - (current_top + offset) + 1
    else
      a = (ts.to_i % p).zero?
      ans[ts.to_sym] = a
      if a
        count += 1
        offset += current_bottom - (current_top + offset) + 1
      end
    end

    break if current_bottom == l_s - 1
    current_bottom += 1
  end

  break if current_top == l_s - l_p
  current_top += 1
end
puts count