#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
using int64 = long long int;

int64 dp[110][100010];
const int64 MOD = 1000000007;
int main() {
    int N, K; cin >> N >> K;
    vector<int> d;
    for(int i=1; i*i<=N; i++) {
        d.push_back(i);
        d.push_back(N/i);
    }
    sort(d.begin(), d.end());
    d.erase(unique(d.begin(), d.end()), d.end());

    vector<int> rng; rng.push_back(1);
    for(size_t i=1; i<d.size(); i++) rng.push_back(d[i] - d[i-1]);

    int M = rng.size();
    dp[0][0] = 1;
    for(int i=0; i<K; i++) {
        for(int j=0; j<M; j++) {
            (dp[i][j+1] += dp[i][j]) %= MOD;
        }
        for(int j=0; j<M; j++) {
            (dp[i+1][j] += dp[i][M-1-j] * rng[j]) %= MOD;
        }
    }

    int64 ans = 0;
    for(int i=0; i<M; i++) (ans += dp[K][i]) %= MOD;
    cout << ans << endl;
    return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#include <algorithm>
using namespace std;
using int64 = long long int;

int64 dp[110][100010];
const int64 MOD = 1000000007;
int main() {
    int N, K; cin >> N >> K;
    vector<int> d;
    for(int i=1; i*i<=N; i++) {
        d.push_back(i);
        d.push_back(N/i);
    }
    sort(d.begin(), d.end());
    d.erase(unique(d.begin(), d.end()), d.end());

    vector<int> rng; rng.push_back(1);
    for(size_t i=1; i<d.size(); i++) rng.push_back(d[i] - d[i-1]);

    int M = rng.size();
    dp[0][0] = 1;
    for(int i=0; i<K; i++) {
        for(int j=0; j<M; j++) {
            (dp[i][j+1] += dp[i][j]) %= MOD;
        }
        for(int j=0; j<M; j++) {
            (dp[i+1][j] += dp[i][M-1-j] * rng[j]) %= MOD;
        }
    }

    int64 ans = 0;
    for(int i=0; i<M; i++) (ans += dp[K][i]) %= MOD;
    cout << ans << endl;
    return 0;
}

```