#include<bits/stdc++.h>
using namespace std;
const int maxn=3e5+10,mod=1e9+7;
int a,b,f[maxn],k,s[maxn],dp2[maxn];
map<int,int,greater<int>> dp;
void plusle(int &a,int b){
    a+=b;if(a>=mod)a-=mod; return;
}
void minun(int &a,int b){
    a-=b; if(a<0)a+=mod; return;
}
int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
int mul(int a,int b){
    return (int)(((long long)a*b)%mod);
}

int main(){
    int n;
    scanf("%d%d",&n,&k);
    int cc=n,pt=1;
    while(cc){
        int x=n/cc;
        dp[x]=(n/x-n/(x+1));
        cc=(n/(x+1));
    }
    int sum=0;
    int last=-1;
    pt=0;
    for(auto it:dp){
        s[pt]=it.second;
        f[pt]=it.second;
        pt++;
    }
    while(k--){
        last=-1;
    for(int i=0;i<pt;i++){
        dp2[i]=(last==-1)?s[i]:add(last,s[i]);
        last=dp2[i];
    }
    for(int i=0;i<pt;i++){
        s[i]=mul(f[i],dp2[pt-i-1]);
    }
    }
    printf("%d\n",s[0]);
}

### Prompt

In Cpp, your task is to solve 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;
const int maxn=3e5+10,mod=1e9+7;
int a,b,f[maxn],k,s[maxn],dp2[maxn];
map<int,int,greater<int>> dp;
void plusle(int &a,int b){
    a+=b;if(a>=mod)a-=mod; return;
}
void minun(int &a,int b){
    a-=b; if(a<0)a+=mod; return;
}
int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
int mul(int a,int b){
    return (int)(((long long)a*b)%mod);
}

int main(){
    int n;
    scanf("%d%d",&n,&k);
    int cc=n,pt=1;
    while(cc){
        int x=n/cc;
        dp[x]=(n/x-n/(x+1));
        cc=(n/(x+1));
    }
    int sum=0;
    int last=-1;
    pt=0;
    for(auto it:dp){
        s[pt]=it.second;
        f[pt]=it.second;
        pt++;
    }
    while(k--){
        last=-1;
    for(int i=0;i<pt;i++){
        dp2[i]=(last==-1)?s[i]:add(last,s[i]);
        last=dp2[i];
    }
    for(int i=0;i<pt;i++){
        s[i]=mul(f[i],dp2[pt-i-1]);
    }
    }
    printf("%d\n",s[0]);
}

```