#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
constexpr int MOD = 1e9 + 7;

long dp[101][101010];

int main() {
    cin.tie(0); ios_base::sync_with_stdio(false);
    int N, K; cin >> N >> K;
    vector<int> states;
    for(int i = 0; i * i <= N; ++i) {
        states.push_back(i);
        if(i != 0) states.push_back(N / i);
    }
    sort(states.begin(), states.end());
    int M = states.size();
    
    dp[0][0] = 1;
    for(int i = 1; i <= K; ++i) {
        for(int j = 1; j <= M; ++j) dp[i-1][j] += dp[i-1][j-1];
        for(int j = 0; j <  M; ++j) {
            dp[i][j] = dp[i-1][M - 2 - j] * (states[j+1] - states[j]) % MOD;
        }
    }

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

### Prompt

Please create a solution in cpp 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 <algorithm>
#include <vector>
using namespace std;
constexpr int MOD = 1e9 + 7;

long dp[101][101010];

int main() {
    cin.tie(0); ios_base::sync_with_stdio(false);
    int N, K; cin >> N >> K;
    vector<int> states;
    for(int i = 0; i * i <= N; ++i) {
        states.push_back(i);
        if(i != 0) states.push_back(N / i);
    }
    sort(states.begin(), states.end());
    int M = states.size();
    
    dp[0][0] = 1;
    for(int i = 1; i <= K; ++i) {
        for(int j = 1; j <= M; ++j) dp[i-1][j] += dp[i-1][j-1];
        for(int j = 0; j <  M; ++j) {
            dp[i][j] = dp[i-1][M - 2 - j] * (states[j+1] - states[j]) % MOD;
        }
    }

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

```