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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define int long long
typedef long long ll;
using namespace std;

const ll MAXN = 200000;
const ll INF = 1001001001;
const ll MOD = 1000000007;

ll N, P, ans=0, tmp1=0, tmp2=0;
string S;
vector<ll> A;

signed main(){
    cin >> N >> P >> S;
    A.resize(P,0);
    A[0]++;
    tmp1 = 1;
    for(int i=0;i<N;i++) {
        ll num = S[N-1-i]-'0';
        num *= tmp1;
        tmp2 += num;
        tmp2 %= P;
        A[tmp2]++;
        tmp1 *= 10;
        tmp1 %= P;
    }
    for(int i=0;i<P;i++) {
        ans += A[i]*(A[i]-1)/2;
    }
    cout << ans << endl;
    return 0;
}