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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

n,p=map(int,input().split())
l=list(input())
for i in range(n):
    l[i]=int(l[i])
w=0
wl=[]
po=pow(10,p-2,p)
for i in range(n):
    wl.append(w)
    #if w==0:
        #ans+=rr[n-1-i]
    w-=l[n-1-i]
    w*=po
    w%=p
wl.reverse()
r=[]
rt=0
for i in range(n):
    rt+=l[n-1-i]*(10**i)
    r.append(rt%p)
r.reverse()
d=[0]*p
ans=0
for i in range(n):
    d[r[i]]+=1
    ans+=d[wl[i]]  
print(ans)