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 PyPy2 (5.6.0) (judge verdict: Accepted)

# -*- coding: utf-8 -*-
import sys
from collections import defaultdict

N,P=map(int, sys.stdin.readline().split())
S=sys.stdin.readline().strip()
S=S[::-1]

ans=0
if P in (2,5):
    len_S=len(S)
    for i,x in enumerate(S):
        x=int(x)
        if x%P==0:
            ans+=(len_S-i)
    print ans
else:
    L=[0] 
    val=0
    for i,x in enumerate(S):
        x=int(x)
        val+=pow(10,i,P)*x
        L.append(val%P)

    C=defaultdict(lambda: 0)    #mod2019の値の出現個数を記録
    for x in L:
        ans+=C[x]   #自分とmod2019の結果が同じ値となる出現個数をansに足す
        C[x]+=1     #自分のmod2019の値を出現個数に+1カウント
    print ans