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<cstdio>
#include<map>
using namespace std;

int n, p, mod[200000];
char s[200001];
map<int, long> ma;

int main() {
    scanf("%d %d", &n, &p);
    scanf("%s", s);
    mod[0] = 1;
    long ans = 0;
    for (int i = 1; i < n; i++) mod[i] = mod[i - 1] * 10 % p;
    for (int i = n - 1, j = 1, sum = 0; i >= 0; i--) {
        sum = (sum + (s[i] - '0') * mod[n - 1 - i]) % p;
        if (sum % p == 0) ans++;
        if (ma.find(sum % p) != ma.end()) ans += ma[sum % p];
        ma[sum % p] += 1;
    }
    printf("%ld\n", ans);
    return 0;
}