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: Accepted)

#include <bits/stdc++.h>

/*#pragma GCC optimize("Ofast")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native")   */


//#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
//using namespace __gnu_pbds;

#define ss second
#define ff first
#define int long long
//#define double long double
#define pb push_back

const int N = 2e6 + 51, MOD = 1e9 + 7, mod = 998244353, INF = 2e18;

mt19937 Flash_new_51(51515);

using vi = vector <int>;
using vp = vector <pair <int, int>>;

int ost[N], sum[N];


signed main() {
    #ifdef Flash
     freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    #endif
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, p;
    string s;
    cin >> n >> p >> s;
    int ans = 0;
    if (p == 2 || p == 5) {
        for (int i = n - 1; i >= 0; i--) {
            if ((s[i] - '0') % p == 0) ans += i + 1;
        }
    }
    else {
        ost[0] = 1;
        int tek = 1;
        for (int i = n - 1; i >= 0; i--) {
            sum[i] = (sum[i + 1] + ((s[i] - '0') * tek) % p) % p;
            tek = (tek * 10) % p;
            ans += ost[sum[i]]++;
        }
    }
    cout << ans;
}