#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;

const int mo = 1e9 + 7;
const int N = 1e5 + 10;

int n, k, ans;
int L[N], R[N], val[N], pos[N], cnt, sum[2][N]; //dp[i][j]前i位, a[i]值为L[i] ~ R[i]

int main() {
	scanf("%d%d", &k, &n);
	for(int l = 1, r; l <= k; l = r + 1) {
		r = k / (k / l);
		L[++ cnt] = l; R[cnt] = r; val[cnt] = k / l;
	}
//	printf("cnt = %d\n", cnt);
	for(int i = 1; i <= cnt; i ++) {
		sum[0][i] = 1;
		pos[i] = upper_bound(L + 1, L + cnt + 1, val[i]) - L - 1;
	}
	int cur = 1;
	for(int i = 1; i <= n; i ++, cur ^= 1) {
		for(int j = 1; j <= cnt; j ++) {
			sum[cur][j] = (sum[cur][j - 1] + sum[cur ^ 1][pos[j]] * 1ll * (R[j] - L[j] + 1) % mo) % mo;
		}
	}
	printf("%d\n", sum[cur ^ 1][cnt]);
	return 0;
}

### Prompt

Please formulate 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 <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;

const int mo = 1e9 + 7;
const int N = 1e5 + 10;

int n, k, ans;
int L[N], R[N], val[N], pos[N], cnt, sum[2][N]; //dp[i][j]前i位, a[i]值为L[i] ~ R[i]

int main() {
	scanf("%d%d", &k, &n);
	for(int l = 1, r; l <= k; l = r + 1) {
		r = k / (k / l);
		L[++ cnt] = l; R[cnt] = r; val[cnt] = k / l;
	}
//	printf("cnt = %d\n", cnt);
	for(int i = 1; i <= cnt; i ++) {
		sum[0][i] = 1;
		pos[i] = upper_bound(L + 1, L + cnt + 1, val[i]) - L - 1;
	}
	int cur = 1;
	for(int i = 1; i <= n; i ++, cur ^= 1) {
		for(int j = 1; j <= cnt; j ++) {
			sum[cur][j] = (sum[cur][j - 1] + sum[cur ^ 1][pos[j]] * 1ll * (R[j] - L[j] + 1) % mo) % mo;
		}
	}
	printf("%d\n", sum[cur ^ 1][cnt]);
	return 0;
}

```