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: Accepted)

// author: @Vitalosa

#include <bits/stdc++.h>

using namespace std;
typedef unsigned long long ull;
typedef long long ll;

template <typename T>
ostream& operator<< (ostream &f, const vector<T> &a){
    for(int i = 0; i < a.size()-1; ++i){ f << a[i] << ' '; }
    f << *(a.end()-1);
    return f;
}

template <typename T>
ostream& operator<< (ostream &f, const vector< vector<T> > &a){
    for(int i = 0; i < a.size()-1; ++i){ f << a[i] << endl; }
    f << *(a.end()-1);
    return f;
}

template <typename T>
istream& operator>> (istream &f, vector<T> &a){
    for(auto &el : a){ f >> el; }
    return f;
}

#define maxN 500500

#define int long long

char arr[maxN];

int cnt[maxN];

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    int n,p;
    cin>>n>>p;
    for (int i=1; i<=n; ++i){
        cin>>arr[i];
    }
    int st=1;
    int sum=0;
    int ans=0;
    if (p==2 || p==5){
        for(int i=1; i<=n; ++i){
            if ((arr[i]-'0')%p==0) ans+=i;
        }
    }else{
        for (int i=n; i>0; --i){
            sum+=st*(arr[i]-'0');
            st*=10;
            st%=p;
            sum%=p;
            ans+=cnt[sum];
            if (sum%p==0) ++ans;
            cnt[sum]++;
        }
    }
    cout<<ans<<"\n";
}