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: Runtime Error)

#include <bits/stdc++.h>

#define rep(X,N) for(ll X = 0; X < (N); X++)
#define PI (acos(-1.0))
#define pback push_back
#define mpair make_pair
#define MODN 1000000007
#define ALL(V) (V).begin(),(V).end()
#define CERR if(false) cerr
#define INT_MAX_HALF (INT_MAX / 2)
#define EPS (1e-10)

using namespace std;

typedef long long ll;

int c2i(char c){
    return c - '0';
}

int main(){

    int n,p;
    cin >> n >> p;

    string s;
    cin >> s;

    assert(p != 2);
    assert(p != 5);

    vector<int> pow10modp(n + 1, 1);
    pow10modp[0] = 0;

    for(int i = 2; i < n + 1; i++){
        pow10modp[i] = pow10modp[i - 1] * 10 % p;
    }

    for(int i = n; i >= 1; i--){
        CERR << pow10modp[i] << " ";
    }
    CERR << endl;

    vector<int> simple_each_digit_mod(n + 1, 0);

    for(int i = 1; i <= n; i++){
        int tmp = c2i(s[n - i]);
        simple_each_digit_mod[i] = tmp % p;
    }

    for(int i = n; i >= 1; i--){
        CERR << simple_each_digit_mod[i] << " ";
    }
    CERR << endl;

    vector<int> each_digit_mod(n + 1, 0);

    for(int i = 1; i <= n; i++){
        each_digit_mod[i] = (each_digit_mod[i - 1] + simple_each_digit_mod[i] * pow10modp[i]) % p;
    }

    for(int i = n; i >= 1; i--){
        CERR << each_digit_mod[i] << " ";
    }
    CERR << endl;


    vector<ll> count(p, 0);

    for(int i = 0; i < n + 1; i++){
        count[each_digit_mod[i]]++;
    }

    ll ans = 0;

    for(int i = 0; i < p; i++){
        ans += count[i] * (count[i] - 1) / 2;
    }

    cout << ans << endl;



    return 0;
}