#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<string>
using namespace std;
int gcd(int a, int b) {
	int c = a % b;
	while (c != 0) {
		a = b;
		b = c;
		c = a % b;
	}
	return b;
}
struct UnionFind {
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

	UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
		for (int i = 0; i < N; i++) par[i] = i;
	}

	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}

	void unite(int x, int y) { // xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
	}

	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};
typedef long long ll;

ll M = 1000000007;

vector<ll> fac(300001); //n!(mod M)
vector<ll> ifac(300001); //k!^{M-2} (mod M)

ll mpow(ll x, ll n) {
	ll ans = 1;
	while (n != 0) {
		if (n & 1) ans = ans * x % M;
		x = x * x % M;
		n = n >> 1;
	}
	return ans;
}
void setcomb() {
	fac[0] = 1;
	ifac[0] = 1;
	for (ll i = 0; i < 300000; i++) {
		fac[i + 1] = fac[i] * (i + 1) % M; // n!(mod M)
		ifac[i + 1] = ifac[i] * mpow(i + 1, M - 2) % M; // k!^{M-2} (mod M) ←累乗にmpowを採用
	}
}
ll comb(ll a, ll b) {
	if (a == 0 && b == 0)return 1;
	if (a < b || a < 0)return 0;
	ll tmp = ifac[a - b] * ifac[b] % M;
	return tmp * fac[a] % M;
}
// mod. m での a の逆元 a^{-1} を計算する
long long modinv(long long a) {
	long long b = M, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= M;
	if (u < 0) u += M;
	return u;
}
vector<vector<ll>> mul(vector<vector<ll>> a, vector<vector<ll>> b, int n) {
	int i, j, k, t;
	vector<vector<ll>> c(n);
	for (i = 0; i < n; i++) {
		for (j = 0; j < n; j++) {
			t = 0;
			for (k = 0; k < n; k++)
				t = (t + a[i][k] * b[k][j] % M) % M;
			c[i].push_back(t);
		}
	}
	return c;
}

template< typename Monoid >
struct SegmentTree {

	int sz;
	vector< Monoid > seg;

	const Monoid M1;

	SegmentTree(int n, const Monoid &M1) : M1(M1) {
		sz = 1;
		while (sz < n) sz <<= 1;
		seg.assign(2 * sz, M1);
	}

	void set(int k, const Monoid &x) {
		seg[k + sz] = x;
	}

	void build() {
		for (int k = sz - 1; k > 0; k--) {
			seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
		}
	}

	void update(int k, const Monoid &x) {
		k += sz;
		seg[k] = x;
		while (k >>= 1) {
			seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
		}
	}

	Monoid query(int a, int b) {
		Monoid L = M1, R = M1;
		for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
			if (a & 1) L = f(L, seg[a++]);
			if (b & 1) R = f(seg[--b], R);
		}
		return f(L, R);
	}

	Monoid value(const int &k) const {
		return seg[k + sz];
	}

	Monoid f(Monoid a, Monoid b) {
		return a + b;
	}
};

int main() {
	ll n, k, i, j, root, dp[101][32000], sum[101][32000], t, ans, sum2, p;
	cin >> n >> k;
	i = 1;
	while (i*i < n)
		i++;
	root = i;
	dp[0][1] = 1;
	sum[1][0] = 0;
	for (j = 1; j <= root; j++)
		sum[0][j] = 1;
	for (j = 1; j <= root; j++) {
		dp[1][j] = 1;
		sum[1][j] = sum[1][j - 1] + 1;
	}
	for (i = 2; i <= k; i++) {
		sum[i][0] = 0;
		sum2 = 0;
		p = root;
		for (j = root; j >= 1; j--) {
			t = n / j;
			if (t > root) {
				dp[i][j] = sum[i - 1][root];
				sum2 += sum[i - 2][j] * (t - p);
				sum2 %= M;
				p = t;
				dp[i][j] += sum2;
				dp[i][j] %= M;
			}
			else
				dp[i][j] = sum[i - 1][t];
		}
		for (j = 1; j <= root; j++)
			sum[i][j] = (sum[i][j - 1] + dp[i][j]) % M;
	}
	ans = sum[k][root];
	p = root;
	for (j = root; j >=1; j--) {
		t = n / j;
		if (t > root) {
			ans += sum[k - 1][j] * (t - p) % M;
			ans %= M;
			p = t;
		}
	}
	cout << ans << endl;
}


Measured memory_bytes: 55168