Codeforces Round #240 (Div. 1)---B.Mashmokh and ACM(dp)

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]表示长度为i,第i个数为j的方案数

/*************************************************************************
    > File Name: cf-240-div1-B.cpp
    > Author: ALex
    > Mail: [email protected]
    > Created Time: 2015年06月02日 星期二 12时35分58秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

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;

static const int mod = 1000000007;
LL dp[2010][2010];

int main() {
    int n, k;
    while (cin >> n >> k) {
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= n; ++i) {
            dp[1][i] = 1;
        }
        for (int i = 1; i < k; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (!dp[i][j]) {
                    continue;
                }
                for (int l = j; l <= n; l +=j) {
                    dp[i + 1][l] += dp[i][j];
                    dp[i + 1][l] %= mod;
                }
            }
        }
        LL ans = 0;
        for (int i = 1; i <= n; ++i) {
            ans += dp[k][i];
            ans %= mod;
        }
        cout << ans << endl;
    }
    return 0;
}