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 Ruby (2.3.3) (judge verdict: Runtime Error)

#include "bits/stdc++.h"
using namespace std;
using ll = long long;
#define rep(i, begin, n) for (int i = begin; i < n; i++)
#define repe(i, begin, n) for (int i = begin; i <= n; i++)
#define repr(i, begin, n) for (int i = begin; i > begin - n; i--)
#define repre(i, begin, end) for (int i = begin; i >= end; i--)

template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}

template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}

const int inf = 1000000007;
const int MOD = 1000000007;
const long long INF = 1000000000000000007;

// -------------------------------------------------------
ll N, P;
string S;
ll sub[201010];

int main()
{
    cin >> N >> P;
    cin >> S;
    if (P == 2)
    {
        ll ans = 0;
        rep(i, 0, N)
        {
            if ((S[i] - '0') % 2 == 0)
            {
                ans += i + 1;
            }
        }
        cout << ans << endl;
        return 0;
    }
    if (P == 5)
    {
        ll ans = 0;
        rep(i, 0, N)
        {
            if (S[i] == '5' || S[i] == '0')
            {
                ans += i + 1;
            }
        }
        cout << ans << endl;
        return 0;
    }

    ll tenfactor = 1;
    vector<ll> cnt(P, 0);
    cnt[0]++;
    for (ll i = N - 1; i >= 0; i--)
    {
        ll num = S[i] - '0';
        sub[i] = (num * tenfactor + sub[i + 1]) % P;
        cnt[sub[i]]++;
        tenfactor = (tenfactor * 10) % P;
    }
    ll ans = 0;
    for (int i = 0; i < P; i++)
    {
        ll ci = cnt[i];
        if (ci >= 2)
        {
            ans += ci * (ci - 1) / 2;
        }
    }
    cout << ans << endl;
}