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 Rust (1.15.1) (judge verdict: Runtime Error)

use std::io::{stdout, BufReader, BufWriter, Read, Write};

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = Read::bytes(BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}
#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};
    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}
#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };
    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };
    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };
    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };
    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };
    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    let out = stdout();
    let mut writer = BufWriter::new(out.lock());
    solve(&mut writer);
    writer.flush().unwrap();
}

fn solve<W: Write>(mut writer: W) {
    #[allow(unused_macros)]
    macro_rules! print {
        ($fmt:expr) => {
            write!(writer, $fmt).unwrap()
        };
        ($fmt:expr, $($arg:tt)*) => {
            write!(writer, $fmt, $($arg)*).unwrap()
        };
    }
    #[allow(unused_macros)]
    macro_rules! println {
        ($fmt:expr) => {
            writeln!(writer, $fmt).unwrap()
        };
        ($fmt:expr, $($arg:tt)*) => {
            writeln!(writer, $fmt, $($arg)*).unwrap()
        };
    }

    // Main Code
    input! { n:usize, p:usize, s:chars }
    let mut rem = vec![0; n];
    let mut ans = 0;
    for i in 0..n {
        let d = s[i].to_string().parse::<usize>().unwrap();
        let mut tmp = vec![0; n];
        for j in 0..p {
            tmp[(10 * j + d) % p] += rem[j];
        }
        tmp[d % p] += 1;
        ans += tmp[0];
        rem = tmp;
    }
    println!("{}", ans);
}