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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const ll maxs = 1e6+5;
const ll lmaxs = 20;
ll mod = 1e9+7;
ll oo = 1e9;

#define pb push_back
#define mp make_pair
#define ff first
#define ss second


int main(){
    freopen("input00.txt","r",stdin);
    ll T = 1;
    //cin>>T;
    while(T--){
        string s;
        ll  n ,p;
        cin>>n>>p;
        cin>>s;

        if(p==2){
            ll ANS=0;
            for(ll i=0;i<s.size();i++){
                if((s[i]-'0')%2==0 )
                    ANS += i+1;
            }
            cout << ANS;
            continue;
        }

        if(p==5){
            ll ANS=0;
            for(ll i=0;i<s.size();i++){
                if((s[i]-'0')%5==0){
                    ANS += i+1;
                }
            }
            cout << ANS ;
            continue;
        }
        ll ANS=0;
        ll N = 0;
        map<ll,ll> MAP;
        MAP[0]++;
        ll mf=1;
        for(ll i=0;i<n;i++){    
            N = (N*10 + (s[i] - '0') )%p;
            ANS += MAP[N];
            MAP[N]++;
            mf  = (mf*10)%p;
        }

        cout << ANS;
    }

    return 0;
}