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: Accepted)

N, P = gets.chomp.split(" ").map(&:to_i)
S = gets.chomp

if P == 2 || P == 5
  count = 0
  (N - 1).downto(0) do |i|
    count += i + 1 if S[i].to_i % P == 0
  end
  puts count 
else
  mods = Array.new(P, 0)
  sum = 0
  bit = 1
  (N - 1).downto(0) do |i|
    sum += S[i].to_i * bit
    sum %= P
    mods[sum] += 1
    bit = bit * 10 % P
  end
  count = mods[0]
  for i in (0..P-1)
    count += mods[i] * (mods[i] - 1) / 2
  end
  puts count
  end