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;

#define db(x) cout << (x) << '\n';
#define all(x) x.begin(),x.end()
#define UNIQUE(x) sort(all(x)),x.resize(unique(all(x))-x.begin())
#define LB(v,x) lower_bound(all(v),x)
#define UB(v,x) upper_bound(all(v),x)
#define endl "\n"
#define F first
#define S second
#define PB push_back
#define PF push_front
#define int long long
#define trick int m = (l+r)>>1,lc=n<<1,rc=lc|1

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vi> vii;

int dy[]={0,1,1,1,0,-1,-1,-1};
int dx[]={-1,-1,0,1,1,1,0,-1};
const int maxn = 1e5;
const int OO = 1e18;
const int mod = 1e9 + 7;
const double PI = acos(-1);

void maxself(int &a,int b)
{a=max(a,b);}
void minself(int &a,int b)
{a=min(a,b);}

int32_t main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

//	freopen("name.in","r",stdin);
//	freopen("name.out","w",stdout);
	
	int n,p;
	string s;
	cin >> n >> p >> s;
	vector<vector<int>>dp(n+1);
	int cont=0;
	for(int i=0;i<n;i++)
	{
		int num = s[i]-'0';
		num%=p;
		for(int j=0;j<dp[i-1].size();j++)
		{
			int ele = dp[i-1][j] + num;
			ele%=p;
			if(ele == 0)
			cont++;
			dp[i].PB(ele);
		}
		if(num==0)
		cont++;
		dp[i].PB(num);


	}
	cout << cont;
	return 0;
}