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.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    static int[][] arr;

    public static void main(String[] args) {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        Scanner sc = new Scanner(br);
        int n = sc.nextInt();
        int k = sc.nextInt();
        arr = new int[n + 2][100];
        for (int i = 1; i <= n + 1; i++) {
            int x = 0;
            for (int j = 1; j <= i; j++) {
                if (i % j == 0) {
                    arr[i][x++] = j;
                }
            }
            arr[i][x++] = 0;
        }
        // System.out.println(new Date().getTime());
        // System.out.println(new Date().getTime());
        Pair dp[][] = new Pair[n + 1][k + 1];
        for (int i = 0; i <= k; i++) {
            dp[1][i] = new Pair(1, 1);
        }
        for (int i = 0; i <= n; i++) {
            dp[i][1] = new Pair(i, 1);
        }
        for (int j = 2; j <= k; j++) {
            for (int i = 2; i <= n; i++) {
                long total = 0;
                int tempArr[] = getAllDevisors(i);
                for (int temp = 0; tempArr[temp] > 0; temp++) {
                    total += dp[tempArr[temp]][j - 1].ansOfEndingTillNow;
                }
                dp[i][j] = new Pair(total + dp[i - 1][j].answer, total);
            }
        }
        System.out.println(dp[n][k].answer);
        // System.out.println(new Date().getTime());
        // for (int i = 1; i <= n; i++) {
        // for (int j = 1; j <= k; j++) {
        // System.out.print(dp[i][j].ansOfEndingTillNow + " ");
        // }
        // System.out.println();
        // for (int j = 1; j <= k; j++) {
        // System.out.print(dp[i][j].answer + " ");
        // }
        // System.out.println();
        // }
    }

    private static int[] getAllDevisors(int i) {
        return arr[i];
    }
}

class Pair {
    long answer;
    long ansOfEndingTillNow;
    static long mod = 1000000007;

    public Pair(long answer, long ansOfEndingTillNow) {
        this.answer = answer % mod;
        this.ansOfEndingTillNow = ansOfEndingTillNow % mod;
    }
}