#include <bits/stdc++.h>
using namespace std;
long long N, K;
const int MOD = 1000000007;
long long dp[101][110000] = {};
long long S[110000] = {};
long long num[110000];

int solve() {
    int M = 0;
    for(int i=1; i<=N;) {
        int j = N / i;
        if(i <= j) {
            num[M++] = 1; 
            ++i;
        } else {
            num[M++] = N / j - i + 1;
            i = N / j + 1;
        }
    }

    dp[0][0] = 1;
    for(int k=0; k<K; ++k) {
        memset(S, 0, sizeof(S));
        for(int j=0; j<M; ++j) S[j + 1] = (S[j] + dp[k][j]) % MOD;
        for(int j=0; j<M; ++j) dp[k + 1][j] = (S[M - j] * num[j]) % MOD;
    }

    int res = 0;
    for(int j=0; j<=M; ++j) res = (res + dp[K][j]) % MOD;
    return res;
}

int main() {
    cin >> N >> K;
    cout << solve() << endl;
}
### 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 <bits/stdc++.h>
using namespace std;
long long N, K;
const int MOD = 1000000007;
long long dp[101][110000] = {};
long long S[110000] = {};
long long num[110000];

int solve() {
    int M = 0;
    for(int i=1; i<=N;) {
        int j = N / i;
        if(i <= j) {
            num[M++] = 1; 
            ++i;
        } else {
            num[M++] = N / j - i + 1;
            i = N / j + 1;
        }
    }

    dp[0][0] = 1;
    for(int k=0; k<K; ++k) {
        memset(S, 0, sizeof(S));
        for(int j=0; j<M; ++j) S[j + 1] = (S[j] + dp[k][j]) % MOD;
        for(int j=0; j<M; ++j) dp[k + 1][j] = (S[M - j] * num[j]) % MOD;
    }

    int res = 0;
    for(int j=0; j<=M; ++j) res = (res + dp[K][j]) % MOD;
    return res;
}

int main() {
    cin >> N >> K;
    cout << solve() << endl;
}
```