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)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(vec) vec.begin(),vec.end()
typedef long long ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
const ll mod=1e9+7;
const int inf=1<<30;

int a[200010];
int s[200010];

int main(){
  ll n,p; cin >> n >> p;
  string t; cin >> t;
  if(p==2){
    ll ans=n*(n+1)/2;
    rep(i,n){
      if((t[i]-'0')%2==1) ans-=n-i;
    }
    cout << ans << endl;
    return 0;
  }
  if(p==5){
    ll ans=n*(n+1)/2;
    rep(i,n){
      if(t[i]!='0'&&t[i]!='5') ans-=n-i;
    }
    cout << ans << endl;
    return 0;
  }
  ll x10=1;
  rep(i,n){
    a[i]=(t[i]-'0')*x10%p;
    x10*=10; x10%=p;
  }
  rep(i,n+1){
    s[i+1]=(s[i]+a[i])%p;
  }
  map<int,int> mp;
  rep(i,n+1){
    mp[s[i]]++;
  }
  ll ans=0;
  for(auto c:mp){
    ans+=c.second*(c.second-1)/2;
  }
  cout << ans << endl;
}