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 <algorithm>
#include <boost/optional.hpp>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <functional>
#include <iostream>
#include <limits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <tuple>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

#define REP(i, n) for(int i = 0, i##_MACRO = (n); i < i##_MACRO; i++)
#define RANGE(i, a, b) for(int i = (a), i##_MACRO = (b); i < i##_MACRO; i++)
#define EACH(e, a) for(auto&& e : a)
#define ALL(a) std::begin(a), std::end(a)
#define RALL(a) std::rbegin(a), std::rend(a)
#define FILL(a, n) memset((a), n, sizeof(a))
#define FILLZ(a) FILL(a, 0)
#define INT(x) (static_cast<int>(x))

using namespace std;

using ll = long long;
using VI = vector<int>;
using VI2D = vector<vector<int>>;

constexpr int INF = 2e9;
constexpr double EPS = 1e-10;
constexpr double PI = acos(-1.0);

constexpr int dx[] = {-1, 0, 1, 0};
constexpr int dy[] = {0, -1, 0, 1};

template <typename T>
constexpr int sign(T x) {
	return x < 0 ? -1 : x > 0 ? 1 : 0;
}

template <>
constexpr int sign(double x) {
	return x < -EPS ? -1 : x > EPS ? 1 : 0;
}

template <typename T, typename U>
constexpr void chmax(T& m, U x) {
	m = max(m, x);
}

template <typename T, typename U>
constexpr void chmin(T& m, U x) {
	m = min(m, x);
}

template <typename T>
constexpr T square(T x) {
	return x * x;
}

// 繰り返し2乗法
// 計算量 O(logn)
template <typename T>
constexpr T pow(T a, int n, int p) {
	T ret = 1;
	while(n != 0) {
		if(n % 2) {
			ret *= a;
			ret %= p;
		}
		a *= a;
		a %= p;
		n /= 2;
	}
	return ret;
}

int main() {
	int n, p;
	string s;
	cin >> n >> p;
	cin >> s;
	if(p == 2 || p == 5) {
		ll result = 0;
		REP(i, n) {
			if(s[i] - '0' == p) {
				result += i + 1;
			}
		}
		cout << result << endl;
		return 0;
	}

	ll result = 0;
	vector<ll> u(n + 1);
	unordered_map<int, int> cnt;
	u[0] = 0;
	cnt[0] = 1;
	REP(i, n) {
		u[i + 1] = u[i] + (s[n - i - 1] - '0') * pow(10, i, p);
		u[i + 1] %= p;
		result += cnt[u[i + 1]];
		cnt[u[i + 1]]++;
		// cout << "u[" << i + 1 << "] = " << u[i + 1] << '\t' << "cnt["
		//	 << u[i + 1] << "] = " << cnt[u[i + 1]] << '\t'
		//	 << "result = " << result << endl;
	}
	cout << result << endl;
	return 0;
}