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.chomp.split.map(&:to_i)
s = gets.chomp.split('').map(&:to_i)

if p == 2
  ans = 0
  1.upto(n) do |i|
    if s[i-1].even?
      ans += i
    end
  end
  p ans
  exit
end

if p == 5
  ans = 0
  1.upto(n) do |i|
    if s[i-1] % 5 == 0
      ans += i
    end
  end
  p ans
  exit
end

s.reverse!
r = []
s.each_with_index do |ss, i|
  if i == 0
    r[i] = ss % p
  end
  am = (ss * (10**i)) % p
  r[i] = (r[i-1] + am) % p
end

ans = 0
ans += r.count(0)

0.upto(10000) do |i|
  c = r.count(i)
  ans += (c * (c-1)) / 2
end
p r
p ans