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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
#include <stack>
#include <iomanip>
using namespace std;
#define all(x) x.begin(), x.end()
typedef vector <int> vi;
typedef pair<int,int> ii;
typedef long long ll;
typedef long double ld;
const int mod = 1e9 + 7;
const ll inf = 3e18 + 5;
int add(int a, int b) { return (a += b) < mod ? a : a - mod; }
int mul(int a, int b) { return 1LL * a * b % mod; }
int sub(int a, int b) { return (a -= b) < 0 ? a + mod : a; }
int ctz(int x) { return __builtin_ctz(x); }
int clz(int x) { return __builtin_clz(x); }

int main(){
  ios_base::sync_with_stdio(false); cin.tie(0);
  #ifdef LOCAL
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
  #endif

  int n, p;
  cin >> n >> p;
  string s;
  cin >> s;
  int sum = 0;
  map<int, ll> mp;
  for(char c : s){
    sum *= 10;
    sum += (c - '0');
    sum %= p;
    mp[sum]++;
  }
  ll res = 0;
  int off = 0;
  for(char c : s){
    res += mp[off];
    off *= 10;
    off += (c - '0');
    off %= p;
    mp[off]--;
  }
  cout << res;
}