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>
using namespace std;
const int Maxn = 200005;
typedef long long ll;
char S[Maxn];
ll N; 
ll P;
ll ans[Maxn];
ll MayAns[Maxn];
ll gs;
ll ANS; 
map<ll,ll> Dict; 
int main()
{
	cin>>N>>P;
	cin>>S+1;
	gs = 1;
	MayAns[0] = 1;
	Dict[0] = 1;
	ans[N+1]=0;
	for(int i=N;i>=1;i--)
	{
		ll Lost = 0;
		for(int j=i;j<=N;j++)
		{
			Lost = Lost *  10 + S[j] - '0';
			Lost %= P;
		}
		ans[i] = Lost;

		if(Dict[Lost] == 0)
		{
			gs++;
			MayAns[gs] = Lost;
			Dict[Lost]++;
		}
		else
			Dict[Lost]++;
	}
	
	for(int i=1;i<=gs;i++)
	{
		if(Dict[MayAns[i]]<2)	continue;
		ANS += Dict[MayAns[i]] *(Dict[MayAns[i]]-1)/2;
	}
	cout<<ANS;
	return 0;
}