#include<cstdio>
using namespace std;
const long long MOD = 1000000000 + 7;
long long dp[100000][110];
int main() {

    int n, k;
    scanf("%d %d", &n, &k);

    int last = n;
    int tmp = 0, ss = 0;
    for(int i = 1 ; i*i <= n ; i++) {
        dp[i][0] = last - n/(i+1), last = n/(i+1);
        tmp=i;
        ss += dp[i][0];
    }
    while(n - ss > 0) {
        dp[++tmp][0] = 1;
        ss++;
    }

    // for(int i = 1 ; i <= tmp ; i++) {
    //     printf("%d\n", dp[i][0]);
    // }

    for(int j = 1 ; j < k ; j++) {
        long long sum = 0;
        for(int i = 1 ; i <= tmp ; i++) {
            sum += dp[tmp-i+1][j-1];
            sum %= MOD;
            dp[i][j] = sum * dp[i][0] % MOD;
        }
    }
    long long ans = 0;
    for(int i = 1 ; i <= tmp ; i++){
        ans += dp[i][k-1] % MOD;
    }
    printf("%lld\n", ans % MOD);

    return 0;
}
### Prompt

Generate 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<cstdio>
using namespace std;
const long long MOD = 1000000000 + 7;
long long dp[100000][110];
int main() {

    int n, k;
    scanf("%d %d", &n, &k);

    int last = n;
    int tmp = 0, ss = 0;
    for(int i = 1 ; i*i <= n ; i++) {
        dp[i][0] = last - n/(i+1), last = n/(i+1);
        tmp=i;
        ss += dp[i][0];
    }
    while(n - ss > 0) {
        dp[++tmp][0] = 1;
        ss++;
    }

    // for(int i = 1 ; i <= tmp ; i++) {
    //     printf("%d\n", dp[i][0]);
    // }

    for(int j = 1 ; j < k ; j++) {
        long long sum = 0;
        for(int i = 1 ; i <= tmp ; i++) {
            sum += dp[tmp-i+1][j-1];
            sum %= MOD;
            dp[i][j] = sum * dp[i][0] % MOD;
        }
    }
    long long ans = 0;
    for(int i = 1 ; i <= tmp ; i++){
        ans += dp[i][k-1] % MOD;
    }
    printf("%lld\n", ans % MOD);

    return 0;
}
```