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;
using ll = long long;
using ld = long double;
using VI = vector<int>;
using VL = vector<ll>;
using loop = int;
template<class T> using PQ = priority_queue<T, vector<T>, greater<T>>;
#define FOR(i,a,n) for(loop (i)=(a);(i)<(n);++(i))
#define eFOR(i,a,n) for(loop (i)=(a);(i)<=(n);++(i))
#define rFOR(i,a,n) for(loop (i)=(n)-1;(i)>=(a);--(i))
#define erFOR(i,a,n) for(loop (i)=(n);(i)>=(a);--(i))
#define each(i, a) for(auto &i : a)
#define SORT(i) sort((i).begin(),(i).end())
#define rSORT(i,a) sort((i).begin(),(i).end(),(a))
#define all(i) (i).begin(),(i).end()
#define out(y,x) ((y) < 0 || h <= (y) || (x) < 0 || w <= (x))
#define line cout << "------------------------\n" 
#define ENDL(i,n) ((i) == (n) - 1 ? "\n" : " ")
#define stop system("pause") //comment out this on AOJ.
constexpr ll INF = 1000000000;
constexpr ll LLINF = 1LL << 60;
constexpr ll mod = 1000000007;
constexpr ll MOD = 998244353;
constexpr ld eps = 1e-10;
constexpr ld pi = 3.1415926535897932;
template<class T>inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; }return false; }
template<class T>inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; }return false; }
inline void init() { cin.tie(nullptr); cout.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); }
template<class T>inline istream& operator>>(istream& is, vector<T>& v) { for (auto& elemnt : v)is >> elemnt; return is; }
template<class T, class U>inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template<class T>inline vector<T> vec(size_t a) { return vector<T>(a); }
template<class T>inline vector<T> defvec(T def, size_t a) { return vector<T>(a, def); }
template<class T, class... Ts>inline auto vec(size_t a, Ts... ts) { return vector<decltype(vec<T>(ts...))>(a, vec<T>(ts...)); }
template<class T, class... Ts>inline auto defvec(T def, size_t a, Ts... ts) { return vector<decltype(defvec<T>(def, ts...))>(a, defvec<T>(def, ts...)); }

int main() {
    init();

    ll n, p; string s; cin >> n >> p >> s;
    VL a(n + 1);
    FOR(i, 0, n) {
        if (0 < i)(a[i + 1] += a[i] * 10) %= p;
        (a[i + 1] += s[i] - '0') %= p;
    }

    map<int, ll> ma;
    ++ma[0];
    ll ans = 0;
    eFOR(i, 1, n) {
        ans += ma[a[i]];
        ++ma[a[i]];
    }
    cout << ans << "\n";
}