#include <bits/stdc++.h>

#define rep(i, n) for (int i = 0; i < int(n); i++)
#define rrep(i, n) for (int i = int(n) - 1; i >= 0; i--)
#define reps(i, n) for (int i = 1; i <= int(n); i++)
#define rreps(i, n) for (int i = int(n); i >= 1; i--)
#define repc(i, n) for (int i = 0; i <= int(n); i++)
#define rrepc(i, n) for (int i = int(n); i >= 0; i--)
#define repi(i, a, b) for (int i = int(a); i < int(b); i++)
#define repic(i, a, b) for (int i = int(a); i <= int(b); i++)
#define repp(i, n) rep(i, n)
#define reppp(i, n) repp(i, n)
#define rrepp(i, n) rrep(i, n)
#define repss(i, n) reps(i, n)
#define repcc(i, n) repc(i, n)
#define repii(i, a, b) repi(i, a, b)
#define each(x, y) for (auto &x : y)
#define all(a) (a).begin(), (a).end()
#define bit(b) (1ll << (b))

using namespace std;

using i32 = int;
using i64 = long long;
using u64 = unsigned long long;
using f80 = long double;
using vi32 = vector<i32>;
using vi64 = vector<i64>;
using vu64 = vector<u64>;
using vf80 = vector<f80>;
using vstr = vector<string>;

inline void yes() { cout << "Yes" << '\n'; exit(0); }
inline void no() { cout << "No" << '\n'; exit(0); }
inline i64 gcd(i64 a, i64 b) { if (min(a, b) == 0) return max(a, b); if (a % b == 0) return b; return gcd(b, a % b); }
inline i64 lcm(i64 a, i64 b) { return a / gcd(a, b) * b; }
inline u64 xorshift() { static u64 x = 88172645463325252ull; x = x ^ (x << 7); return x = x ^ (x >> 9); }
void solve(); int main() { ios::sync_with_stdio(0); cin.tie(0); cout << fixed << setprecision(16); solve(); return 0; }
template <typename T> class pqasc : public priority_queue<T, vector<T>, greater<T>> {};
template <typename T> class pqdesc : public priority_queue<T, vector<T>, less<T>> {};
template <typename T> inline void amax(T &x, T y) { if (x < y) x = y; }
template <typename T> inline void amin(T &x, T y) { if (x > y) x = y; }
template <typename T> inline T power(T x, i64 n, T e = 1) { T r = e; while (n > 0) { if (n & 1) r *= x; x *= x; n >>= 1; } return r; }
template <typename T> istream& operator>>(istream &is, vector<T> &v) { each(x, v) is >> x; return is; }
template <typename T> ostream& operator<<(ostream &os, vector<T> &v) { rep(i, v.size()) { if (i) os << ' '; os << v[i]; } return os; }
template <typename T, typename S> istream& operator>>(istream &is, pair<T, S> &p) { is >> p.first >> p.second; return is; }
template <typename T, typename S> ostream& operator<<(ostream &os, pair<T, S> &p) { os << p.first << ' ' << p.second; return os; }

template <int mod>
struct ModInt {
  int x;
  ModInt(): x(0) {}
  ModInt(i64 a) { x = a % mod; if (x < 0) x += mod; }
  ModInt &operator+=(ModInt that) { x = (x + that.x) % mod; return *this; }
  ModInt &operator-=(ModInt that) { x = (x + mod - that.x) % mod; return *this; }
  ModInt &operator*=(ModInt that) { x = (i64) x * that.x % mod; return *this; }
  ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
  ModInt operator-() { return ModInt(-this->x); }
  friend ostream& operator<<(ostream &os, ModInt m) { return os << m.x; }
  ModInt inverse() {
    int a = x, b = mod, u = 1, v = 0;
    while (b) { int t = a / b; a -= t * b; u -= t * v; swap(a, b); swap(u, v); }
    return ModInt(u);
  }
  #define op(o, p) ModInt operator o(ModInt that) { return ModInt(*this) p that; }
    op(+, +=) op(-, -=) op(*, *=) op(/, /=)
  #undef op
  #define op(o) bool operator o(ModInt that) const { return x o that.x; }
    op(==) op(!=) op(<) op(<=) op(>) op(>=)
  #undef op
};

using mint = ModInt<1000000007>;

void solve() {
  int N, K; cin >> N >> K;

  vi32 t; {
    set<int> se = { 0 };
    for (int i = 1; i * i <= N; i++) {
      se.insert(i), se.insert(N / i);
    }
    each(s, se) t.push_back(s);
  }
  int M = t.size() - 1;
  
  vector<vector<mint>> dp(2, vector<mint>(M + 1));
  vector<vector<mint>> ac(2, vector<mint>(M + 1));
  reps(j, M) ac[0][j] = 1;

  rep(i, K) {
    int p = (i + 0) % 2;
    int n = (i + 1) % 2;
    reps(j, M) {
      int o = lower_bound(all(t), N / t[j]) - t.begin();
      /*
       * 今回選ぶ整数は、区間 (t[j - 1], t[j]] にある整数.
       * 区間の大きさ R = t[j] - t[j - 1]
       * 区間内の整数を x_1, x_2, ..., x_R とおく.
       * すると N / x_1 = N / x_2 = ... = N / x_R が成り立つ.
       * したがって、Σ_{j=1}^{N/x_R} dp[p][j] 通りの選び方がある.
       */
      dp[n][j] += ac[p][o] * (t[j] - t[j - 1]);

      // dpの累積和を更新する
      ac[n][j] += ac[n][j - 1] + dp[n][j];
    }
    rep(j, M + 1) dp[p][j] = 0;
    rep(j, M + 1) ac[p][j] = 0;
  }
  cout << ac[K & 1][M] << '\n';
}


Measured memory_bytes: 3584