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>
#include <unordered_map>
#include <unordered_set>
using namespace std;
#define endl "\n"
#define ll long long
#define oo 0x3f3f3f3fLL
#define sz(s) (int)(s.size())
#define RT(s) return cout<<s,0
#define INF 0x3f3f3f3f3f3f3f3fLL
#define all(v) v.begin(),v.end()
#define watch(x) cout<<(#x)<<" = "<<x<<endl
const int dr[] { -1, -1, 0, 1, 1, 1, 0, -1 };
const int dc[] { 0, 1, 1, 1, 0, -1, -1, -1 };
#if __cplusplus >= 201402L
template<typename T>
vector<T> create(size_t n) {
	return vector<T>(n);
}
template<typename T, typename ... Args>
auto create(size_t n, Args ... args) {
	return vector<decltype(create<T>(args...))>(n, create<T>(args...));
}
#endif
void run() {
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
#ifndef ONLINE_JUDGE
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#else
#endif
}

ll power(ll x, ll y, int mod) {
	if (y == 0)
		return 1;
	if (y == 1)
		return x % mod;
	ll r = power(x, y >> 1, mod);
	return (((r * r) % mod) * power(x, y & 1, mod)) % mod;
}

ll modInverse(ll b, ll mod) {
	return power(b, mod - 2, mod);
}

int main() {
	//run();
	int n, p;
	string s;
	cin >> n >> p >> s;
	for (auto &it : s)
		it -= '0';
	ll ans = 0;
	vector<ll> frq(p);
	frq[0]++;
	ll val = 0;
	for (int i = 1; i <= n; i++) {
		val = ((val * 10) + s[i - 1]) % p;
		ll tmp = val;
		tmp *= power(10, 1LL * i * (p - 2), p);
		tmp %= p;
		ans += frq[tmp];
		frq[tmp]++;
	}
	cout << ans;
}