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.awt.Point;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.*;
import java.util.*;

public class Main {
	//////
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	//////
	public static long modularExponentiation(long x, long n, long M) {
		long result = 1;
		while (n > 0) {
			if (n % 2 == 1)
				result = (result * x) % M;
			x = (x * x) % M;
			n = n / 2;
		}
		return result;
	}

	public static long modularExponentiation1(long x, long n) {
		long result = 1;
		while (n > 0) {
			if (n % 2 == 1)
				result = (result * x);
			x = (x * x);
			n = n / 2;
		}
		return result;
	}

	/////

	static public int fun(int num) {
		int prod = 1;
		while (num > 0) {
			if (num % 10 != 0) {
				prod = prod * (num % 10);
			}
			num = num / 10;
		}
		return prod;
	}

	public static boolean dfs(int x, int find, boolean[] visited, int[][] matrix) {

		boolean ans = false;
		if (x == find) {
			ans = true;
			return true;
		}
		for (int i = 0; i < 26; i++) {
			if (visited[i] == false && matrix[x][i] == 1) {
				visited[i] = true;
				ans = ans || dfs(i, find, visited, matrix);
			}
		}

		return ans;
	}

	static int lis(int arr[], int n, int k, int b) {
		int lis[] = new int[n];
		int i, j, max = 0;

		/* Initialize LIS values for all indexes */
		for (i = 0; i < n; i++)
			lis[i] = 1;

		/* Compute optimized LIS values in bottom up manner */
		for (i = 1; i < n; i++)
			for (j = 0; j < i; j++)
				if (arr[i] >= k * arr[j] + b && lis[i] < lis[j] + 1)
					lis[i] = lis[j] + 1;

		/* Pick maximum of all LIS values */
		for (i = 0; i < n; i++)
			if (max < lis[i])
				max = lis[i];

		return max;
	}

	public static void main(String[] args) throws IOException {

		FastReader scan = new FastReader();
		PrintWriter pw = new PrintWriter(System.out);

		int Nused = scan.nextInt();
		int length = scan.nextInt();
		long MOD=1000000007;
		long[][] dp = new long[length + 1][Nused + 1];
		for(int i=1;i<=Nused;i++){
			dp[1][i]=1;
		}
		for (int i = 2; i <= length; i++) {
			for (int k = 1; k <= Nused; k++) {
				for (int j = 1; j*k <= Nused; j ++) {
					dp[i][k] += dp[i - 1][j*k]%MOD;}
			}
		}
		long sum=0;
		for(int i=1;i<=Nused;i++){
			sum+=dp[length][i]%MOD;
			sum=sum%MOD;
		}
		//if(sum<0){
		//	sum+=MOD;
		//}
		System.out.println(sum);
		//System.out.println(dp[length][Nused]);

	}
}