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>
    using namespace std;
    using ll = long long;
    #define pb push_back
    #define eb emplace_back
    #define mp make_pair
    #define ff first
    #define ss second

    void io(){
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);
        #ifdef LOCAL
            freopen("input.txt","r",stdin);
            freopen("output.txt","w",stdout);
        #endif
    }
    ll ans;
    int x;
    const int N = 2e5 + 5;
    int main() {
        io();
        string s;
        int n, p;
        cin >> n >> p;
        cin >> s;
        map <int, int> f;
        if(p == 2) {
            for(int i = 0; i < n; i++) {
                if((s[i] - '0') % 2 == 0) {
                    ans += (i + 1);
                }
            }
            cout << ans;
        }
        else if(p == 5) {
            for(int i = 0; i < n; i++) {
                if(s[i] == '5' || s[i] == '0') {
                    ans += (i + 1);
                }
            }
            cout << ans;
        }
        else {
            reverse(s.begin(), s.end());
            int num = 0;
            f[0] = 1;
            int t = 1;
            for(int i = 0; i < n; i++) {
                num = num * t + (s[i] - '0');
                t = t * 10 % p;
                num = num % p;
                f[num]++;


            }        

            for(auto it : f) {

                    ans += ((1LL * it.second * (it.second - 1)) / 2);

            }
            cout << ans;
        }



    }