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>
#define SORT(x) sort((x).begin(), (x).end())
#define ALL(x) (x).begin(), (x).end()
#define rep(i, n) for (ll i = 0; i < n; i++)
#define reps(i, m, n) for (ll i = m; i < n; i++)
#define repr(i, m, n) for (ll i = m; i >= n; i--)
#define de(x) cout << #x << "=" << x << endl;
#define SP << " " <<
template <class T> bool maxi(T &a, const T &b) {
    if (a < b) { a = b; return 1;} return 0;}
template <class T> bool mini(T &a, const T &b) {
    if (b < a) { a = b; return 1;} return 0;}
#define dame {cout<< "-1" <<"\n"; return;}
#define INF2 1000000000000000037
#define INF  1000000007
#define MOD  1000000007
using namespace std;
using ll = long long;
using ld = long double;
//using P  = pair<ll,ll>;
//--GLOBAL---------------------------------

//--MAIN-----------------------------------
void Main() {
    ll N,P; cin>>N>>P;
    string S; cin>>S;
    reverse(ALL(S));
    ll zen[N+1]={};
    ll now=1;
    rep(i,N){
        zen[i+1]=zen[i]+now*(S[i]-'0');
        zen[i+1]%=P;
        now=(now*10%P)+P;
    }

    ll ans=0;
    map<ll,ll> mp;
    reps(i,1,N+1){
        ++mp[zen[i-1]];
        ans+=mp[zen[i]];
    }
    cout<< ans <<"\n";

}

//--START----------------------------------
int main() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    Main();
}
//-----------------------------------------