#include <iostream>
#include <vector>
using namespace std;

int main() {
    long long const mod = 1e9+7;
    int n, k, rn = 1;
    vector<int> v, op;
    vector< vector<long long> > dp;
    cin >> n >> k;
    while(rn*rn <= n){
        v.push_back(rn);
        rn++;
    }
    if(rn-1 != n/(rn-1))v.push_back(n/(rn-1));
    for(int i=rn-3;i>=0;i--){
        v.push_back(n/v[i]);
    }
    int vs = v.size();
    op.assign(vs, 0);
    for(int i=0;i<vs;i++){
        op[i] = vs-1-i;
    }
    dp.assign(2, vector<long long>(vs));

    for(int i=0;i<vs;i++){
        dp[0][i] = v[i];
    }
    for(int i=1;i<k;i++){
        int ni = i%2;
        int bi = 1-i%2;
        dp[ni][0] = dp[bi][vs-1];
        for(int j=1;j<vs;j++){
            long long res = dp[bi][op[j]];
            res *= (v[j] - v[j-1]);
            res += dp[ni][j-1];
            dp[ni][j] = res % mod;
        }
    }
    cout << dp[(k+1)%2][vs-1] << endl;
	return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;

int main() {
    long long const mod = 1e9+7;
    int n, k, rn = 1;
    vector<int> v, op;
    vector< vector<long long> > dp;
    cin >> n >> k;
    while(rn*rn <= n){
        v.push_back(rn);
        rn++;
    }
    if(rn-1 != n/(rn-1))v.push_back(n/(rn-1));
    for(int i=rn-3;i>=0;i--){
        v.push_back(n/v[i]);
    }
    int vs = v.size();
    op.assign(vs, 0);
    for(int i=0;i<vs;i++){
        op[i] = vs-1-i;
    }
    dp.assign(2, vector<long long>(vs));

    for(int i=0;i<vs;i++){
        dp[0][i] = v[i];
    }
    for(int i=1;i<k;i++){
        int ni = i%2;
        int bi = 1-i%2;
        dp[ni][0] = dp[bi][vs-1];
        for(int j=1;j<vs;j++){
            long long res = dp[bi][op[j]];
            res *= (v[j] - v[j-1]);
            res += dp[ni][j-1];
            dp[ni][j] = res % mod;
        }
    }
    cout << dp[(k+1)%2][vs-1] << endl;
	return 0;
}

```