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: Accepted)

#pragma region template
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vld = vector<ld>;
using vvld = vector<vld>;
using vvvld = vector<vvld>;
using vs = vector<string>;
using pll = pair<ll, ll>;
using vp = vector<pll>;
template <typename T>
using pqrev = priority_queue<T, vector<T>, greater<T>>;
#define rep(i, n) for (ll i = 0, i##_end = (n); i < i##_end; i++)
#define repb(i, n) for (ll i = (n)-1; i >= 0; i--)
#define repr(i, a, b) for (ll i = (a), i##_end = (b); i < i##_end; i++)
#define reprb(i, a, b) for (ll i = (b)-1, i##_end = (a); i >= i##_end; i--)
#define ALL(a) (a).begin(), (a).end()
#define SZ(x) ((ll)(x).size())
//*
constexpr ll MOD = 1e9 + 7;
/*/
constexpr ll MOD = 998244353;
//*/
constexpr ll INF = 1e+18;
constexpr ld EPS = 1e-12L;
constexpr ld PI = 3.14159265358979323846L;
constexpr ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
constexpr ll LCM(ll a, ll b) { return a / GCD(a, b) * b; }
template <typename S, typename T>
inline bool chmax(S &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename S, typename T>
inline bool chmin(S &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
#ifdef OJ_LOCAL
#include "dump.hpp"
#else
#define dump(...) ((void)0)
#endif
template <typename T>
bool print_(const T &a) {
  cout << a;
  return true;
}
template <typename T>
bool print_(const vector<T> &vec) {
  for (auto &a : vec) {
    cout << a;
    if (&a != &vec.back()) {
      cout << " ";
    }
  }
  return false;
}
template <typename T>
bool print_(const vector<vector<T>> &vv) {
  for (auto &v : vv) {
    for (auto &a : v) {
      cout << a;
      if (&a != &v.back()) {
        cout << " ";
      }
    }
    if (&v != &vv.back()) {
      cout << "\n";
    }
  }
  return false;
}
void print() { cout << "\n"; }
template <typename Head, typename... Tail>
void print(Head &&head, Tail &&... tail) {
  bool f = print_(head);
  if (sizeof...(tail) != 0) {
    cout << (f ? " " : "\n");
  }
  print(forward<Tail>(tail)...);
}
#pragma endregion

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  ll n, p;
  cin >> n >> p;
  string s_;
  cin >> s_;
  vll s(n);
  rep(i, n) s[i] = s_[i] - '0';
  ll ans = 0;
  if(p == 2){
    rep(i, n){
      if(s[i]%2 == 0){
        ans += i+1;
      }
    }
  }else if(p == 5){
    rep(i, n){
      if(s[i]%5 == 0){
        ans += i+1;
      }
    }
  }else{
    map<ll, ll> cnt;
    cnt[0]++;
    ll c = 1;
    ll cur = 0;
    repb(i, n){
      cur += c * s[i];
      cur %= p;
      cnt[cur]++;
      c *= 10;
      c %= p;
    }
    dump(cnt);
    for(auto&& e: cnt){
      ans += e.second*(e.second-1)/2;
    }
  }
  print(ans);
}