#include <bits/stdc++.h>
using namespace std;
#define rep(i,s,t) for(int i=s;i<t;i++)
const int mod=1e9+7;
void add(int &a,int b)
{
    a+=b;
    if(a>=mod)a-=mod;
}
void sub(int &a,int b)
{
    a-=b;
    if(a<0)a+=mod;
}
int allnum,n,k,dp[111][2][33323],w[33323];
int main()
{
    scanf("%d%d",&k,&n);
    allnum=sqrt(k);
    dp[1][0][0]=dp[1][1][0]=1;
    rep(j,1,allnum+1)w[j]=k/j-max(allnum,k/(j+1));
    rep(i,1,n+1)
    {
        rep(j, 1, allnum + 1)
        {
            add(dp[i][0][j], dp[i][0][j - 1]), add(dp[i][1][j], dp[i][1][j - 1]);
            add(dp[i + 1][0][0], dp[i][0][j]);
            add(dp[i + 1][1][j], dp[i][0][j]);
        }
        rep(j, 1, allnum + 1)
        {
            dp[i][1][j]=dp[i][1][j]*1ll*w[j]%mod;
            add(dp[i + 1][0][0], dp[i][1][j]), sub(dp[i + 1][0][j + 1], dp[i][1][j]);
        }
    }
    int res=0;
    rep(x,0,2)rep(j,1,allnum+1)
        add(res,dp[n][x][j]);
    printf("%d\n",res);
}
### 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 <bits/stdc++.h>
using namespace std;
#define rep(i,s,t) for(int i=s;i<t;i++)
const int mod=1e9+7;
void add(int &a,int b)
{
    a+=b;
    if(a>=mod)a-=mod;
}
void sub(int &a,int b)
{
    a-=b;
    if(a<0)a+=mod;
}
int allnum,n,k,dp[111][2][33323],w[33323];
int main()
{
    scanf("%d%d",&k,&n);
    allnum=sqrt(k);
    dp[1][0][0]=dp[1][1][0]=1;
    rep(j,1,allnum+1)w[j]=k/j-max(allnum,k/(j+1));
    rep(i,1,n+1)
    {
        rep(j, 1, allnum + 1)
        {
            add(dp[i][0][j], dp[i][0][j - 1]), add(dp[i][1][j], dp[i][1][j - 1]);
            add(dp[i + 1][0][0], dp[i][0][j]);
            add(dp[i + 1][1][j], dp[i][0][j]);
        }
        rep(j, 1, allnum + 1)
        {
            dp[i][1][j]=dp[i][1][j]*1ll*w[j]%mod;
            add(dp[i + 1][0][0], dp[i][1][j]), sub(dp[i + 1][0][j + 1], dp[i][1][j]);
        }
    }
    int res=0;
    rep(x,0,2)rep(j,1,allnum+1)
        add(res,dp[n][x][j]);
    printf("%d\n",res);
}
```