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.util.*;
public class Abc {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		int ar[][] = new int[n + 1][k + 1];
		long fact[][] = new long[2001][2001];
		long mod = 1000000007;
		for(int i = 0; i < fact.length; i++) {
			for(int j = 0; j < fact.length; j++) {
				if(i == 0 || j == 0) {
					fact[i][j] = 1;
				} else if(i == 1) {
					fact[i][j] = 1;
				} else if(j == 1) {
					fact[i][j] = i;
				} else {
					fact[i][j] = (fact[i - 1][j] + fact[i][j - 1]) % mod;
				}
			}
		}
		ar[n][1] = 1;
		for(int i = n - 1; i >= 1; i --) {
			ar[i][1] = 1;
			for(int j = 2 * i; j <= n; j += i) {
				for(int l = 2; l <= k && ar[j][l - 1] > 0; l++) {
					ar[i][l] += ar[j][l - 1]; 
				}
			}
		}

		long sum[] = new long[k + 1];
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= k; j++) {
				sum[j] += ar[i][j];
			}
		}
		long s = 0;
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= k; j ++) {
				s = (s + (ar[i][j] * fact[j][k - j] % mod)) % mod;
				//System.out.println("fact["+ j +"][" + (k - j) + "] = "+ fact[j][k - j]);
			}
		}
		System.out.println(s);
	}

	public static long pow(long a, long b, long mod) {
		long result = 1;
		while(b > 0) {
			if(b % 2 == 1) {
				result = (result * a) % mod;
			}
			a = (a * a) % mod;
			b /= 2;
		}
		return result;
	}
}