#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
#include <cmath>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const ll MOD = 1e9 + 7;
ll n, k;

vector <pll> ans;

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

	cin >> n >> k;

	ll i = 1;
	for (; i*i <= n; i++)
	{
		ans.push_back({ 1, 1 });
	}

	ll last = --i;
	if (i*i == n) i--;
	for (i; i > 0; i--)
	{
		ans.push_back({ n / i - last, 1 });
		last += n / i - last;
	}
	if (last < n)
		ans.push_back({n-last,1});

	for (int i = 0; i < k; i++)
	{
		ll psum[100000] = { 0, };
		for (int j = 0; j < ans.size(); j++)
		{
			psum[j] = ans[j].first * ans[j].second % MOD;
			if (j) psum[j] += psum[j - 1], psum[j] %= MOD;
		}
		for (int j = 0; j < ans.size(); j++)
		{
			ans[j].second = psum[ans.size() - j - 1];
		}
	}

	cout << ans[0].second;
}
### Prompt

Generate a CPP solution to the following problem:

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.

Examples

Input

3 2


Output

5


Input

10 3


Output

147


Input

314159265 35


Output

457397712

### Response

```cpp
#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
#include <cmath>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const ll MOD = 1e9 + 7;
ll n, k;

vector <pll> ans;

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

	cin >> n >> k;

	ll i = 1;
	for (; i*i <= n; i++)
	{
		ans.push_back({ 1, 1 });
	}

	ll last = --i;
	if (i*i == n) i--;
	for (i; i > 0; i--)
	{
		ans.push_back({ n / i - last, 1 });
		last += n / i - last;
	}
	if (last < n)
		ans.push_back({n-last,1});

	for (int i = 0; i < k; i++)
	{
		ll psum[100000] = { 0, };
		for (int j = 0; j < ans.size(); j++)
		{
			psum[j] = ans[j].first * ans[j].second % MOD;
			if (j) psum[j] += psum[j - 1], psum[j] %= MOD;
		}
		for (int j = 0; j < ans.size(); j++)
		{
			ans[j].second = psum[ans.size() - j - 1];
		}
	}

	cout << ans[0].second;
}
```