Mashmokh’s boss, Bimokh, didn’t like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh’s team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn’t able to solve them. That’s why he asked you to help him with these tasks. One of these tasks is the following.

A sequence of l integers b1, b2, …, bl (1 ≤ b1 ≤ b2 ≤ … ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally for all i (1 ≤ i ≤ l - 1).

Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).

Input

The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).

Output

Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).

Sample test(s)

Input

3 2

Output

5

Input

6 4

Output

39

Input

2 1

Output

2

Note

In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].

dp[i][j] 长度为i,第i个数为j的方案数

/*************************************************************************

    > File Name: CF240D.cpp

    > Author: ALex

    > Mail: zchao1995@gmail.com

    > Created Time: 2015年03月17日 星期二 11时59分48秒

 ************************************************************************/

#include <map>

#include <set>

#include <queue>

#include <stack>

#include <vector>

#include <cmath>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const double pi = acos(-1.0);

const int inf = 0x3f3f3f3f;

const double eps = 1e-15;

typedef long long LL;

typedef pair <int, int> PLL;

const int mod = 1000000007;

LL dp[2010][2010];

int main ()

{

    int n, K;

    while(~scanf("%d%d", &n, &K))

    {

        memset (dp, 0, sizeof(dp));

        for (int i = 1; i <= n; ++i)

        {

            dp[1][i] = 1;

        }

        for (int i = 2; i <= K; ++i)

        {

            for (int j = 1; j <= n; ++j)

            {

                for (int k = 1; k * k <= j; ++k)

                {

                    if (j % k == 0)

                    {

                        dp[i][j] += dp[i - 1][k] + (k * k == j ? 0 : dp[i - 1][j / k]);

                        dp[i][j] %= mod;

                    }

                }

            }

        }

        LL ans = 0;

        for (int i = 1; i <= n; ++i)

        {

            ans += dp[K][i];

            ans %= mod;

        }

        printf("%lld\n", ans);

    }

    return 0;

}