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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

int main(){
    string S;
    int N;
    int P;
    int tmp = 0;
    long long int count = 0;
    cin >> N >> P;
    cin >> S;
    if(P==2 || P==5){
        for(int j=0;j<N;j++){
            if((S[j] - '0')%P==0)count += (j+1);
    }}
    else{
    int dp[P+10];
    int u = 1;
    for(int j=0;j<P;j++)dp[j]=0;
    for(int j=N-1;j>=0;j--){
        tmp = (S[j] - '0')*u + tmp;
        tmp %= P;
        dp[tmp]++;
        u = (u*10)%P;
    }
    for(int j=1;j<P;j++)count += dp[j]*(dp[j]-1)/2;
    count += dp[0]*(dp[0]+1)/2;
    }
    cout << count << endl;
}