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 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
int main()
{
	long long int n, p;
	scanf("%lld %lld", &n, &p);
	char s[200005];
	scanf("%s", s);
	long long int i, j;
	long long int dp[2][10004];
	for (i = 0; i < p; i++)
		dp[0][i] = 0;
	long long int ans = 0;
	for (i = 0; i < n; i++)
	{
		for (j = 0; j < p; j++)
			dp[1][j] = 0;
		for (j = 0; j < p; j++)
			dp[1][(10 * j + s[i] - '0') % p] += dp[0][j];
		dp[1][(s[i] - '0') % p]++;
		for (j = 0; j < p; j++)
			dp[0][j] = dp[1][j];
		ans += dp[0][0];
	}
	printf("%lld\n", ans);
	return 0;
}