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 Go (1.6) (judge verdict: Time Limit Exceeded)

package main

import (
	"fmt"
	"math"
	"math/big"
)

func main() {
	var n, p int64
	var s string
	fmt.Scanf("%d %d", &n, &p)
	fmt.Scanf("%s", &s)
	answer := countDivisible(n, p, s)
	fmt.Println(answer)
}

func countDivisible(n, p int64, s string) int64 {
	modMatrix := make([][]int64, len(s))

	for i := 0; i < len(s); i++ {
		modMatrix[i] = make([]int64, len(s))
		for j := 0; j < len(s)-i; j++ {
			// i: index in s, j: 10 ** j
			if j < 14 { // rough estimation for avoiding int overflow
				modMatrix[i][j] = (int64(s[i]) - 48) * int64(math.Pow10(j)) % p
			} else {
				bigP := big.NewInt(p)
				bigSi := big.NewInt(int64(s[i]) - 48)
				ten := big.NewInt(10)
				ten.Exp(ten, big.NewInt(int64(j)), nil)
				bigSi.Mul(bigSi, ten)
				bigSi.Mod(bigSi, bigP)
				modMatrix[i][j] = bigSi.Int64()
			}
		}
	}

	var cnt int64
	for i := 0; i < len(s); i++ {
		for j := i; j < len(s); j++ {
			var sumOfMod int64
			for k := i; k <= j; k++ {
				sumOfMod += modMatrix[k][j-k]
			}
			if sumOfMod%p == 0 {
				cnt++
			}
		}
	}
	return cnt
}