Problem:
Score : 600 points

Problem Statement

Find the number of sequences of length K consisting of positive integers such that the product of any two adjacent elements is at most N, modulo 10^9+7.

Constraints

1\leq N\leq 10^9

1 2\leq K\leq 100 (fixed at 21:33 JST)

N and K are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of sequences, modulo 10^9+7.

Sample Input 1

3 2

Sample Output 1

5

(1,1), (1,2), (1,3), (2,1), and (3,1) satisfy the condition.

Sample Input 2

10 3

Sample Output 2

147

Sample Input 3

314159265 35

Sample Output 3

457397712

Solution in Rust (1.15.1) (judge verdict: Accepted)

#[allow(dead_code)]
fn is_whitespace(c: &char) -> bool {
    c.is_whitespace()
}

#[allow(dead_code)]
fn is_eol(c: &char) -> bool {
    ['\r', '\n'].contains(c)
}

#[allow(unused_macros)]
macro_rules! input {
    ($var:ident, $($r:tt)*) => {
        input_inner!{$var, $($r)*}
    };

    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        #[allow(unused_mut, unused_variables, bare_trait_objects)]
        let mut next = move |f: &Fn(&char)->bool| -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|f(&c))
                .take_while(|c|!f(&c))
                .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 ; eol ]) => {
        $next(&is_eol).split(' ').map(|x| x.parse::<$t>().unwrap()).collect::<Vec<_>>()
    };

    ($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, parser) => {
        $next
    };

    ($next:expr, $t:ty) => {
        $next(&is_whitespace).parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{BinaryHeap, HashMap};

#[allow(dead_code)]
type Graph = Vec<Vec<usize>>;
#[allow(dead_code)]
const M: usize = 1e9 as usize + 7;

fn main() {
    input! {
        n: usize,
        k: usize,
    }

    let mut v = std::collections::HashSet::new();
    for i in 1..n+1 {
        if i * i > n {
            break;
        }
        v.insert(i);
        v.insert(n / i);
    }
    let mut v = v.into_iter().collect::<Vec<_>>();
    v.sort();

    let mut limits = vec![0; v.len()];
    for i in 0..v.len() {
        let x = v.binary_search(&(n/v[i])).unwrap();
        limits[i] = x;
    }

    let mut cand = vec![0; v.len()];
    cand[0] = 1;
    for i in 1..v.len() {
        cand[i] = v[i] - v[i-1];
    }

    // println!("{:?}", v);
    // println!("{:?}", limits);
    // println!("{:?}", cand);

    let mut dp = vec![vec![0; v.len()]; k + 1];
    dp[0][0] = 1;
    for i in 1..k+1 {
        let mut sum = vec![0; v.len() + 1];
        sum[0] = dp[i-1][0];
        for j in 1..v.len() {
            sum[j] = sum[j-1] + dp[i-1][j];
            sum[j] %= M;
        }

        for j in 0..v.len() {
            let mut next = sum[limits[j]];
            next *= cand[j];
            next %= M;
            dp[i][j] = next;
        }
    }
    // println!("{:?}", dp);
    let mut ans = 0;
    for i in 0..v.len() {
        ans += dp[k][i];
        ans %= M;
    }
    println!("{}", ans);
}