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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<algorithm>
#include<climits>
#include<cmath>
#include<cstring>
#include<deque>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<utility>
#include<vector>

using namespace std;

using lint = long long;
using P = pair<int, int>;
using LLP = pair<long long, long long>;

#define REP(i, x, n) for(int i = (x), i##_len = int(n) ; i < i##_len ; ++i)
#define rep(i, n) for(int i = 0, i##_len = int(n) ; i < i##_len ; ++i)
#define reps(i, n) for(int i = 1, i##_len = int(n) ; i <= i##_len ; ++i)
#define rrep(i, n) for(int i = int(n) - 1 ; i >= 0 ; --i)
#define rreps(i, n) for(int i = int(n) ; i > 0 ; --i)
#define SORT(x) sort((x).begin(), (x).end())
#define SORT_INV(x) sort((x).rbegin(), (x).rend())
#define TWINS(x) cout << ((x) ? "Yay!" : ":(") << endl

const int IINF = (1 << 30) - 1;
const long long LLINF = 1LL << 61;
const int dx4[] = {1, 0, -1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy8[] = {0, -1, -1, -1, 0, 1, 1, 1};
const double EPS = 1e-8;

template<long long modulus>
class ModInt{
    long long num;

    long long mod(long long n){
        if(n < 0LL){
            n += (abs(n + 1LL) / modulus + 1LL) * modulus;
        }
        return n % modulus;
    }

    long long ext_gcd(long long a, long long b, long long &x, long long &y){
        if(b == 0LL){
            x = 1;
            y = 0;
            return a;
        }

        long long d = ext_gcd(b, a % b, y, x);
        y -= a / b * x;

        return d;
    }

    long long get_inv(long long n){
        long long x, y;
        ext_gcd(n, modulus, x, y);
        return mod(x);
    }

public:
    ModInt(long long n = 0LL){
        init(n);
    }

    void init(long long n = 0LL){
        num = mod(n);
        return;
    }

    long long get(void){
        return num;
    }

    ModInt &operator+=(const ModInt &rhs){
        num += rhs.num;
        if(num >= modulus){
            num -= modulus;
        }
        return *this;
    }

    ModInt &operator-=(const ModInt &rhs){
        num -= rhs.num;
        if(num < 0LL){
            num += modulus;
        }
        return *this;
    }

    ModInt &operator*=(const ModInt &rhs){
        num = num * rhs.num % modulus;
        return *this;
    }

    ModInt &operator/=(const ModInt &rhs){
        long long inv = get_inv(rhs.num);
        num = num * inv % modulus;
        return *this;
    }

    ModInt operator+(const ModInt &rhs) const {
        return ModInt(*this) += rhs;
    }

    ModInt operator-(const ModInt &rhs) const {
        return ModInt(*this) -= rhs;
    }

    ModInt operator*(const ModInt &rhs) const {
        return ModInt(*this) *= rhs;
    }

    ModInt operator/(const ModInt &rhs) const {
        return ModInt(*this) /= rhs;
    }

    ModInt &operator=(const ModInt &rhs){
        num = rhs.num;
        return *this;
    }

    bool operator<(const ModInt &rhs) const {
        return num < rhs.num;
    }
};

template<long long modulus>
istream &operator>>(istream &lhs, ModInt<modulus> &rhs){
    long long val;
    lhs >> val;
    rhs.init(val);
    return lhs;
}

template<long long modulus>
ostream &operator<<(ostream &lhs, ModInt<modulus> &rhs){
    lhs << rhs.get();
    return lhs;
}

using mint = ModInt<1000000007LL>;

lint n;
int k;
vector<lint> v;
mint dp[101][300000];

// lint solve(int i, int j){
//     if(j == 0){
//         return i == k;
//     }else if(dp[i][j] >= 0LL){
//         return dp[i][j];
//     }

//     lint res = solve(i, j - 1);
//     if(i < k){
//         res += solve(i + 1, (int)v.size() + (~j) + 1) * (v[j] - v[j - 1]) % MOD;
//         res %= MOD;
//     }

//     return dp[i][j] = res;
// }

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> n >> k;

    v.emplace_back(0LL);
    for(lint i = 1 ; i * i <= n ; ++i){
        v.emplace_back(i);
        v.emplace_back(n / i);
    }
    SORT(v);

    dp[k][0] = mint(1);
    for(int i = k ; i >= 0 ; --i){
        for(int j = 1 ; j < (int)v.size() ; ++j){
            dp[i][j] += dp[i][j - 1];
            if(i < k){
                dp[i][j] += dp[i + 1][(int)v.size() + (~j) + 1] * mint(v[j] - v[j - 1]);
            }
        }
    }

    cout << dp[0][(int)v.size() - 1] << endl;

    return 0;
}