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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
//#include <atcoder/all>
using namespace std;
#define all(n) begin(n), end(n)
const long long INF = numeric_limits<long long>::max();
typedef long long ll;
typedef vector<int> vint;
typedef vector<vector<int>> vvint;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef unsigned long long ull;
template <class T>
bool chmax(T &a, const T &b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T &a, const T &b)
{
    if (b < a)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
vector<T> make_v(size_t a) { return vector<T>(a); }

template <typename T, typename... Ts>
auto make_v(size_t a, Ts... ts)
{
    return vector<decltype(make_v<T>(ts...))>(a, make_v<T>(ts...));
}
template <typename T, typename V>
typename enable_if<is_class<T>::value == 0>::type
fill_v(T &t, const V &v) { t = v; }

template <typename T, typename V>
typename enable_if<is_class<T>::value != 0>::type
fill_v(T &t, const V &v)
{
    for (auto &e : t)
        fill_v(e, v);
}
const int MAX_N = 1e6 + 1;
vector<long long> factorialMemo(MAX_N, -1);        //factorialMemo[i] := i! mod p
vector<long long> inverseFactorialMemo(MAX_N, -1); // inverseFactorialMemo[i] := (factorialMemo[i])^(-1)
bool initFinished = false;
template <long long mod>
struct ModInt
{
    long long x;

    ModInt() : x(0) {}

    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

    ModInt &operator+=(const ModInt &p)
    {
        if ((x += p.x) >= mod)
            x -= mod;
        return *this;
    }

    ModInt &operator-=(const ModInt &p)
    {
        if ((x += mod - p.x) >= mod)
            x -= mod;
        return *this;
    }

    ModInt &operator*=(const ModInt &p)
    {
        x = (1LL * x * p.x % mod);
        return *this;
    }

    ModInt &operator/=(const ModInt &p)
    {
        *this *= p.inverse();
        return *this;
    }

    ModInt operator-() const { return ModInt(-x); }

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

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

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

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

    bool operator==(const ModInt &p) const { return x == p.x; }

    bool operator!=(const ModInt &p) const { return x != p.x; }

    ModInt inverse() const
    {
        long long a = x, b = mod, u = 1, v = 0, t;
        while (b > 0)
        {
            t = a / b;
            swap(a -= t * b, b);
            swap(u -= t * v, v);
        }
        return ModInt(u);
    }

    ModInt pow(int64_t n) const
    {
        ModInt ret(1), mul(x);
        while (n > 0)
        {
            if (n & 1)
                ret *= mul;
            mul *= mul;
            n >>= 1;
        }
        return ret;
    }

    friend ostream &operator<<(ostream &os, const ModInt &p)
    {
        return os << p.x;
    }

    friend istream &operator>>(istream &is, ModInt &a)
    {
        int64_t t;
        is >> t;
        a = ModInt<mod>(t);
        return (is);
    }

    static long long get_mod() { return mod; }
    static void factorialInit()
    {
        factorialMemo[0] = 1;
        for (ll i = 0; i < MAX_N - 1; i++)
        {
            factorialMemo[i + 1] = ModInt<mod>(factorialMemo[i] * (i + 1)).x;
        }
        initFinished = true;
    }
    static ModInt factorial(ll n)
    {
        if (!initFinished)
        {
            factorialInit();
        }
        return factorialMemo[n];
    }

    static ModInt Combination(ll n, ll k)
    {
        if (n < k)
            return 0;
        return factorial(n) / (factorial(n - k) * factorial(k));
    }
};
using mint = ModInt<(int)1e9 + 7>;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int N, K;
    cin >> N >> K;
    set<int> s;
    for (size_t i = 1; i * i < N; i++)
    {
        s.insert(i);
        s.insert(N / i);
    }
    vint ktov;
    map<int, int> vtok;
    for (auto &&i : s)
    {
        ktov.push_back(i);
    }
    for (size_t i = 0; i < ktov.size(); i++)
    {
        vtok[ktov[i]] = i;
    }
    auto DP = make_v<mint>(K + 1, ktov.size());
    map<int, mint> c;
    for (auto &&i : ktov)
    {
            c[i] = mint(N/i - N/(i+1));
    }
    vector<mint> acc(ktov.size() + 1), tmp = acc;
    DP[0][vtok[N]]=1;

    partial_sum(all(DP[0]), ++acc.begin());
    for (int i = 0; i < K; i++)
    {
        for (auto &&p : tmp)
        {
            p = 0;
        }
        
        for (int j = 0; j < ktov.size(); j++)
        {
            DP[i + 1][j] = c[ktov[j]] * (acc.back() - acc[vtok[N / ktov[j]]]);
            tmp[j + 1] = tmp[j] + DP[i + 1][j];
        }
        acc = tmp;
    }
    mint ans = 0;
    for (auto &&i : DP.back())
    {
        ans += i;
    }
    cout << ans;
    return 0;
}