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)

/*
    Mbak Sana Dong
    Mbak Chaeyoung Dong
*/
#include<bits/stdc++.h>
#define ll long long int
using namespace std;
ll dp[200200];
int main()
{
    int n,i,j,k,x,y,ans,temp,curr,m;
    int tc;
    ll chaeng = 0;
    int p;
    cin >> n >> p;
    string s;
    cin >> s;
    s = " " + s;
    ll sana = 0;
    ll now = 0;
    temp = 0;
    for(i=1;i<=n;i++)
    {
        curr = s[i] - '0';
        now += curr;
        now %= p;
        if(now == 0)
        {
            dp[i] = 1;
            temp++;
        }
        now *= 10;
    }
    chaeng = 0;
    now = 0;
    for(i=1;i<=n;i++)
    {
        chaeng += temp - now*2;
        if(dp[i] == 1)
        {
            now++;
        }
    }
    cout << chaeng << endl;
    return 0;
}