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;
#ifdef DEBUG
#define FPEN freopen("in.txt", "r", stdin);
#define debug(...)                                                             \
    { cerr << "# " << __VA_ARGS__ << "\n"; }
#else
#define debug(...)
#define FPEN
#endif
const int maxn = 200009;
typedef long long i64;
typedef pair<i64, i64> pi64;
typedef pair<int, int> pi;
const i64 MOD = 119 << 23 | 1;
class {
  public:
    i64 c[maxn];
    i64 n, p;
    string s;
    void solve() {
        cin >> n >> p >> s;
        i64 a = 0, ans = 0, deg = 1;
        // if (p == 2 || p == 5) {
        //     for (int i = 0; i < n; ++i) {
        //         if ((s[i] - '0') % p == 0) {
        //             ans += i + 1;
        //         }
        //     }
        //     cout << ans << '\n';
        //     return;
        // }
        c[0] = 1;
        for (int i = 0; i < n; ++i) {
            a = (a * 10 + s[i] - '0') % p;
            if (deg == 0) {
                ans += c[a];
                c[0]++;
            } else {
                ans += c[a * pw(deg, p - 2) % p];
                c[a * pw(deg, p - 2) % p]++;
            }
            deg = deg * 10 % p;
        }
        cout << ans << '\n';
    }
    i64 pw(i64 a, i64 b) {
        i64 res = 1;
        while (b) {
            if (b & 1ll)
                res = res * a % p;
            a = a * a % p;
            b >>= 1;
        }
        return res;
    }
} NSPACE;
int main() {
    FPEN;
    ios_base::sync_with_stdio(false);
    cout.tie(0);
    cin.tie(0);
    NSPACE.solve();
}