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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <array>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <list>
#include <numeric>
#include <stack>
#include <iomanip>
#include <random>
#include <complex>
#include <functional>
#include <tuple>

using namespace std;

#define Rep(i,a,b) for(ll i = a; i < b; ++i)
#define rep(i,b) Rep(i,0,b)
#define allof(a) (a).begin(), (a).end()

#define Yes(q) ((q) ? "Yes" : "No")
#define YES(q) ((q) ? "YES" : "NO")
#define Possible(q) ((q) ? "Possible" : "Impossible")
#define POSSIBLE(q) ((q) ? "POSSIBLE" : "IMPOSSIBLE")

using ll = long long;

constexpr int inf = 1e9 + 7;
constexpr ll infll = 1ll << 60ll;
constexpr ll mod = 1e9 + 7;
// 0~3までは右左下上 4~7までは斜め
constexpr int dx[] = { 1, 0, -1, 0, 1, 1, -1, -1 };
constexpr int dy[] = { 0, -1, 0, 1, 1, -1, -1, 1 };

namespace {
    template<typename T> void chmax(T& a, T b) { a = std::max(a, b); }
    template<typename T> void chmin(T& a, T b) { a = std::min(a, b); }
    template<typename T> void chadd(T& a, T b) { a = a + b; }

    void Cout(long long x, const char* end = "\n") { std::cout << x << end; }
    template <typename T> void Cout(const T& x, const char* end = "\n") { std::cout << x << end; }
    template <typename T> void Cout(const std::vector<T>& x, const char* sep = " ", const char* end = "\n") { for (std::size_t i = 0, sz = x.size(); i < sz; i++) { std::cout << x[i] << (i == sz - 1 ? end : sep); } }

    // 標準入出力
    struct inp {
        std::size_t sz;
        inp(std::size_t _sz = 1) : sz(_sz) {}
        template <typename T> operator T () const { T a; std::cin >> a; return a; }
        template <typename T> operator std::vector<T>() const { std::vector<T> a(sz); for (std::size_t i = 0; i < sz; i++) std::cin >> a[i]; return a; }
    };

    // 2次元用の標準入出力
    template<typename T>
    struct inpn {
        std::size_t szi, szj;
        inpn(std::size_t _szi, std::size_t _szj) : szi(_szi), szj(_szj) {}
        operator std::vector<std::vector<T>>() const {
            std::vector<std::vector<T>> a(szi, std::vector<T>(szj));
            for (std::size_t i = 0; i < szi; ++i)
                for (std::size_t j = 0; j < szj; ++j) cin >> a[i][j];
            return a;
        }
    };

    inp inp1; // input one
}

int main() {

    int n, p;
    cin >> n >> p;
    string s = inp1;
    reverse(allof(s));

    // p = 5 として a = 15 がpで割り切れるなら a * 10 = 150 もpで割り切れる
    // ただし，割り切れない場合はあまりも10倍になる
    // 累積和

    ll ans = 0LL;

    if (p == 2 || p == 5) {
        // 下一桁がpで割り切れればよい
        rep(i, n) {
            if ((s[i] - '0') % p == 0) ans += n - i;
        }
        Cout(ans);
        return 0;
    }

    // (a_l - a_r) / 10^r == 0 (mod.p) となればよい
    // 変形すると a_l (mod.p) == a_r (mod.p)
    vector<ll> a(n + 1, 0LL);
    ll ten_mod = 1;
    rep(i, n) {
        a[i + 1] = (a[i] + ten_mod * (s[i] - '0')) % p;
        ten_mod = (ten_mod * 10) % p;
    }
    vector<ll> v(p, 0);
    rep(i, n + 1) {
        ++v[a[i]];
    }
    rep(i, p) {
        ans += (v[i] - 1) * v[i] / 2;
    }

    Cout(ans);

    return 0;
}