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>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < n; ++i)
#pragma region Debug
template <typename T>
void view(const std::vector<T> &v)
{
    for (const auto &e : v)
    {
        std::cout << e << " ";
    }
    std::cout << std::endl;
}
template <typename T>
void view(const std::vector<std::vector<T>> &vv)
{
    for (const auto &v : vv)
    {
        view(v);
    }
}
#pragma endregion

#pragma region mod
int p = 1e9 + 7; //10^9 + 7
struct mint
{
    ll x;
    mint(ll x = 0) : x(x % p) {}
    mint &operator+=(const mint a)
    {
        if ((x += a.x) >= p)
            x -= p;
        return *this;
    }
    mint &operator-=(const mint a)
    {
        if ((x += p - a.x) >= p)
            x -= p;
        return *this;
    }
    mint &operator*=(const mint a)
    {
        (x *= a.x) %= p;
        return *this;
    }

    mint operator+(const mint a) const
    {
        mint res(*this);
        return res += a;
    }
    mint operator-(const mint a) const
    {
        mint res(*this);
        return res -= a;
    }
    mint operator*(const mint a) const
    {
        mint res(*this);
        return res *= a;
    }

    mint pow(ll t) const
    {
        mint r = 1;
        mint a = *this;
        while (true)
        {
            if (t & 1)
            {
                r *= a;
            }
            t >>= 1;
            if (t == 0)
                break;
            a *= a;
        }

        return r;
    }

    mint inv() const
    {
        return pow(p - 2);
    }

    mint &operator/=(const mint a)
    {
        return (*this) *= a.inv();
    }
};

struct comb
{
    vector<mint> fact, invs;
    comb(int n) : fact(n + 1), invs(n + 1)
    {
        assert(n < p);

        fact[0] = 1;
        for (int i = 1; i <= n; ++i)
            fact[i] = fact[i - 1] * i;

        invs[n] = fact[n].inv();
        for (int i = n; i >= 1; --i)
            invs[i - 1] = invs[i] * i;
    }
    mint operator()(int n, int k)
    {
        if (k < 0 || k > n)
            return 0;
        return fact[n] * invs[n - k] * invs[k];
    }
};
#pragma endregion

string s;
int r, n;

vector<vector<int>> dp;
int f(int i, mint m)
{
    if (i == n)
    {
        return 0;
    }
    if (dp[i][m.x] != -1)
    {
        return dp[i][m.x];
    }

    int x = s.at(i) - '0';
    int newm = (m + r * x).x;
    int res = (newm == 0) + f(i + 1, (m * 10) + x);

    return dp[i][m.x] = res;
}

int main()
{
    cin >> n >> p;
    cin >> s;
    dp = vector<vector<int>>(n + 5, vector<int>(p + 5, -1));
    r = mint(10).inv().x;
    ll ans = 0;
    rep(i, n)
    {
        ans += f(i, 0);
    }
    cout << ans << endl;

    return 0;
}