#include "bits/stdc++.h"
using namespace std;

int main() {
	long long N, K, SQ;
	static long long DP[110][80000] = {}, X[80000], MOD = 1000000007;
	DP[0][0] = 1;
	cin >> N >> K;
	SQ = sqrt(N);
	for (int i = 0; i < SQ; i++) {
		X[i] = 1;
	}
	for (int i = 0; i < SQ; i++) {
		int P = SQ - i;
		X[i + SQ] = N / P - N / (P + 1);
	}
	if (SQ * (SQ + 1) > N) X[SQ]--;
	SQ *= 2;
	for (int i = 0; i < K; i++) {
		long long Cur[80000] = {};
		Cur[0] = DP[i][0];
		for (int j = 1; j < SQ; j++) {
			Cur[j] = Cur[j - 1] + DP[i][j];
			Cur[j] = Cur[j] % MOD;
		}
		reverse(Cur, Cur + SQ);
		for (int j = 0; j < SQ; j++) {
			DP[i + 1][j] = (Cur[j] * X[j]) % MOD;
		}
	}
	long long ANS = 0;
	for (int i = 0; i < SQ; i++) {
		ANS = (ANS + DP[K][i]) % MOD;
	}
	cout << ANS << endl;
}
### Prompt

In CPP, your task is to solve 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;

int main() {
	long long N, K, SQ;
	static long long DP[110][80000] = {}, X[80000], MOD = 1000000007;
	DP[0][0] = 1;
	cin >> N >> K;
	SQ = sqrt(N);
	for (int i = 0; i < SQ; i++) {
		X[i] = 1;
	}
	for (int i = 0; i < SQ; i++) {
		int P = SQ - i;
		X[i + SQ] = N / P - N / (P + 1);
	}
	if (SQ * (SQ + 1) > N) X[SQ]--;
	SQ *= 2;
	for (int i = 0; i < K; i++) {
		long long Cur[80000] = {};
		Cur[0] = DP[i][0];
		for (int j = 1; j < SQ; j++) {
			Cur[j] = Cur[j - 1] + DP[i][j];
			Cur[j] = Cur[j] % MOD;
		}
		reverse(Cur, Cur + SQ);
		for (int j = 0; j < SQ; j++) {
			DP[i + 1][j] = (Cur[j] * X[j]) % MOD;
		}
	}
	long long ANS = 0;
	for (int i = 0; i < SQ; i++) {
		ANS = (ANS + DP[K][i]) % MOD;
	}
	cout << ANS << endl;
}
```