#include <iostream>
#include<bits/stdc++.h>
#define ll long long
#define   ull unsigned long long 
const int MAX=8e4+9;
const long long mod=1e9+7;
using namespace std;
vector<int>adj[MAX];

ull fact[MAX];
ll dp[MAX][101];
ll pre[MAX][101];
int main(){
	ll n, k;

	cin>>n>>k;
	vector<ll>v;
	v.push_back(0);
	for(int i=1;i*i<=n;i++){
		v.push_back(i);
		if(i*i<n)
			v.push_back(n/i);
	}
//	cout<<1;
	map<ll ,ll>mp;
	sort(v.begin(), v.end());
	for(int i=0;i<v.size();i++){
		mp[v[i]]=i;	
	}
	for(int i=1;i<v.size();i++)
		dp[i][1]=1;
	for(int i=1;i<v.size();i++)
		pre[i][1]=v[i];
	for(int i=2;i<=k;i++){
		for(int j=1;j<v.size();j++){
			int idx=mp[n/v[j]];
			dp[j][i]=((v[j]-v[j-1])*pre[idx][i-1])%mod;
			pre[j][i]=(pre[j-1][i]+dp[j][i])%mod;
		}
	}
	cout<<pre[v.size()-1][k];


}
### Prompt

Your task is to create 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<bits/stdc++.h>
#define ll long long
#define   ull unsigned long long 
const int MAX=8e4+9;
const long long mod=1e9+7;
using namespace std;
vector<int>adj[MAX];

ull fact[MAX];
ll dp[MAX][101];
ll pre[MAX][101];
int main(){
	ll n, k;

	cin>>n>>k;
	vector<ll>v;
	v.push_back(0);
	for(int i=1;i*i<=n;i++){
		v.push_back(i);
		if(i*i<n)
			v.push_back(n/i);
	}
//	cout<<1;
	map<ll ,ll>mp;
	sort(v.begin(), v.end());
	for(int i=0;i<v.size();i++){
		mp[v[i]]=i;	
	}
	for(int i=1;i<v.size();i++)
		dp[i][1]=1;
	for(int i=1;i<v.size();i++)
		pre[i][1]=v[i];
	for(int i=2;i<=k;i++){
		for(int j=1;j<v.size();j++){
			int idx=mp[n/v[j]];
			dp[j][i]=((v[j]-v[j-1])*pre[idx][i-1])%mod;
			pre[j][i]=(pre[j-1][i]+dp[j][i])%mod;
		}
	}
	cout<<pre[v.size()-1][k];


}
```