#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

const int mod=1000000007;
ll dp[2][200000],m[200000];
vector<int> v;
int n,k;
int main(){
    cin>>n>>k;
    for(int i=1;i*i<=n;++i){
        v.pb(i);
        v.pb(n/i);
    }
    sort(v.begin(),v.end());
    v.erase(unique(v.begin(),v.end()),v.end());
    int sz=v.size();
    m[0]=1;dp[0][0]=1;
    FOR(i,1,sz) m[i]=v[i]-v[i-1],dp[0][i]=m[i];
    int nx;
    FOR(i,1,k){
        nx=(i&1);
        ll sum=0;
        memset(dp[nx],0,sizeof(dp[nx]));
        rep(j,sz){
            sum+=dp[nx^1][j];sum%=mod;
            dp[nx][sz-j-1]+=sum*m[sz-j-1]%mod;
            dp[nx][sz-j-1]%=mod;
        }
    }
    ll ans=0;
    rep(i,sz) ans+=dp[nx][i];
    cout<<ans%mod<<endl;
    return 0;
}

### Prompt

Create a solution in Cpp for 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 <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

const int mod=1000000007;
ll dp[2][200000],m[200000];
vector<int> v;
int n,k;
int main(){
    cin>>n>>k;
    for(int i=1;i*i<=n;++i){
        v.pb(i);
        v.pb(n/i);
    }
    sort(v.begin(),v.end());
    v.erase(unique(v.begin(),v.end()),v.end());
    int sz=v.size();
    m[0]=1;dp[0][0]=1;
    FOR(i,1,sz) m[i]=v[i]-v[i-1],dp[0][i]=m[i];
    int nx;
    FOR(i,1,k){
        nx=(i&1);
        ll sum=0;
        memset(dp[nx],0,sizeof(dp[nx]));
        rep(j,sz){
            sum+=dp[nx^1][j];sum%=mod;
            dp[nx][sz-j-1]+=sum*m[sz-j-1]%mod;
            dp[nx][sz-j-1]%=mod;
        }
    }
    ll ans=0;
    rep(i,sz) ans+=dp[nx][i];
    cout<<ans%mod<<endl;
    return 0;
}

```