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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/*BISMILLAH
THE WHITE WOLF
NO DREAM IS TOO BIG AND NO DREAMER IS TOO SMALL*/

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef vector<long long> vll;
typedef vector<int> vi;

#define io ios_base::sync_with_stdio(false)
#define pb push_back
#define eb emplace_back
//#define mod   1000000000
#define PI 2*acos(0.0)
#define dbg(a) cout<<#a<<" ->->->-> "<<a<<"\n"
int dirx[] = {1, -1,0, 0}, diry[] = {0, 0, 1, -1};

ll bigmod(ll x, ll p, int mod)
{
    ll res = 1;
    while(p)
    {
        if(p&1)
            res = (res*x)%mod;
        x = (x*x)%mod;
        p >>= 1;
    }
    return res;
}



//=============================================ASIFAZAD==============================================//


int32_t main()
{
    io;
    string s;
    int n, p;
    cin>>n>>p>>s;
    vi cnt(p), modul;
    int am = 0;
    for(int i =0 ; i< n; i++)
    {
        am = (am * 10%p + (s[i] - '0')%p)%p;
        cnt[am*bigmod(10, n-1-i, p)%p]++;
    }
    cnt[0]++;
    ll ans = 0;
    for(auto x: cnt)
        if(x >= 2)
            ans += x*(x-1)/2;
    cout<<ans;
    return 0;
}