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.Arrays;
import java.util.StringTokenizer;

/**
 * _414B
 * θ(k*n*ln(n)) time
 * θ(k*n) space
 *
 * @author artyom
 */
public class _414B implements Runnable {
    private static final int MOD = 1000000007;
    private BufferedReader in;
    private PrintStream out;
    private StringTokenizer tok;

    private void solve() throws IOException {
        int n = nextInt(), k = nextInt();
        long[][] dp = new long[k][n + 1];
        Arrays.fill(dp[k - 1], 1);
        for (int i = k - 2; i >= 0; i--) {
            for (int j = 1; j <= n; j++) {
                long sum = 0;
                for (int d = 1, m = n / j; d <= m; d++) {
                    sum = (sum + dp[i + 1][j * d]) % MOD;
                }
                dp[i][j] = sum;
            }
        }
        long sum = 0;
        for (int j = 1; j <= n; j++) {
            sum = (sum + dp[0][j]) % MOD;
        }
        out.print(sum);
    }

    //--------------------------------------------------------------
    public static void main(String[] args) {
        new _414B().run();
    }

    @Override
    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = System.out;
            tok = null;
            solve();
            in.close();
        } catch (IOException e) {
            System.exit(0);
        }
    }

    private String nextToken() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    private int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    private long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    private int[] readIntArray(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextInt();
        }
        return arr;
    }

    private static int getMax(int[] a) {
        int max = Integer.MIN_VALUE;
        for (int i = 0, n = a.length; i < n; i++) {
            if (a[i] > max) {
                max = a[i];
            }
        }
        return max;
    }

    private static int getMin(int[] a) {
        int min = Integer.MAX_VALUE;
        for (int i = 0, n = a.length; i < n; i++) {
            if (a[i] < min) {
                min = a[i];
            }
        }
        return min;
    }
}