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>
#define WHOLE(v) (v).begin(), (v).end()
#define REV_WHOLE(v) (v).rbegin(), (v).rend()
using i64 = int64_t;
using namespace std;
template<class F>auto recursive(F f){return[f](auto...a){return f(f,a...);};}
template<class I,class V=typename I::value_type>V sum_up(const I&l,const I&r){V v;for(I i=l;i!=r;i++)v=i==l?*i:v+*i;return v;}
template<class I,class T=iterator_traits<I>>I operator+(I it,int n){for(int i=0;i<n;i++)it++;return it;}
template<class I,class T=iterator_traits<I>>I operator-(I it,int n){for(int i=0;i<n;i++)it--;return it;}
template<class T>using rev_priority_queue=priority_queue<T,vector<T>,greater<T>>;
template<class T>using vector2d=vector<vector<T>>;
struct fixprec{int p;fixprec(int p):p(p){}};
ostream&operator<<(ostream&o,fixprec f){return o<<fixed<<setprecision(f.p);}
void R_YESNO(bool p){cout<<(p?"YES":"NO")<<endl;}
void R_YesNo(bool p){cout<<(p?"Yes":"No")<<endl;}
// O(N + P)
int main() {
    int N, P;
    string S;
    cin >> N >> P >> S;
    i64 ans = 0;
    reverse(WHOLE(S));
    // 下一桁だけで判定する場合
    if(P == 2 || P == 5) {
        for(int i = 0; i < N; i++) {
            if( (S[i] - '0') % P == 0) ans += i + 1;
        }
    } else {
        // 値の後ろに0をつけても 0 mod P には関係ない場合
        vector<i64> v(N), m(P, 0);
        i64 p10 = 1;
        for(int i = 0; i < N; i++) {
            v[i] = ((i ? v[i - 1] : 0) + p10 * (S[i] - '0')) % P;
            m[v[i]]++;
            p10 = (p10 * 10) % P;
        }
        ans += m[0] * (m[0] - 1);
        for(int i = 1; i <= P / 2; i++) {
            ans += m[i] * m[P - i];
        }
    }
    cout << ans << endl;
    return 0;
}