Problem: 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 <image> 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).

Examples

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].

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {
    private static final long MOD = 1000000007;
    
    public static void main(String[] args) throws Exception {
        FastScanner scan = new FastScanner(System.in);
        int N = scan.nextInt();
        int K = scan.nextInt();
        long[][] dp = new long[N + 1][K + 1];
        for (int i = 1; i <= N; ++i) {
            for (int j = 1; j <= K; ++j) {
                dp[i][j] = 1;
            }
        }
        for (int i = 2; i <= N; ++i) {
            for (int j = 1; j < i; ++j) {
                if (i % j != 0) {
                    continue;
                }
                long sum = 0;
                for (int k = 1; k < K; ++k) {
                    sum += dp[j][k];
                    dp[i][k + 1] = (dp[i][k + 1] + sum) % MOD;
                }
            }
        }
        long res = 0;
        for (int i = 1; i <= N; ++i) {
            res += dp[i][K];
        }
        System.out.println(res % MOD);
    }
}

class FastScanner {
    BufferedReader in;
    StringTokenizer tok;

    public FastScanner(InputStream in) {
        this.in = new BufferedReader(new InputStreamReader(in));
        tok = new StringTokenizer("");
    }

    private String tryReadNextLine() {
        try {
            return in.readLine();
        } catch (Exception e) {
            throw new InputMismatchException();
        }
    }

    public String nextToken() {
        while (!tok.hasMoreTokens()) {
            tok = new StringTokenizer(next());
        }
        return tok.nextToken();
    }

    public String next() {
        String newLine = tryReadNextLine();
        if (newLine == null)
            throw new InputMismatchException();
        return newLine;
    }

    public int nextInt() {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() {
        return Long.parseLong(nextToken());
    }

}