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.Arrays;
import java.io.DataInputStream;
import java.io.PrintWriter;
import java.io.FileInputStream;
import java.io.IOException;

public class DivisionintoTeams
{
	static class Reader
	{
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;

		public Reader()
		{
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public Reader(String file_name) throws IOException
		{
			din = new DataInputStream(new FileInputStream(file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public String readLine() throws IOException
		{
			byte[] buf = new byte[201]; // line length
			int cnt = 0, c;
			while ((c = read()) != -1)
			{
				if (c == '\n')
					break;
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, cnt);
		}

		public int nextInt() throws IOException
		{
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do
			{
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}

		public long nextLong() throws IOException
		{
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			}
			while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}

		public double nextDouble() throws IOException
		{
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			}
			while ((c = read()) >= '0' && c <= '9');

			if (c == '.')
			{
				while ((c = read()) >= '0' && c <= '9')
				{
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}

		private void fillBuffer() throws IOException
		{
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}

		private byte read() throws IOException
		{
			if (bufferPointer == bytesRead)
				fillBuffer();
			return buffer[bufferPointer++];
		}

		public void close() throws IOException
		{
			if (din == null)
				return;
			din.close();
		}
	}
        
        public static void main(String[] args) throws IOException
	{
		Reader s=new Reader();PrintWriter pw=new PrintWriter(System.out);
                int n=s.nextInt(),k=s.nextInt();
                long dp[][][]=new long[k+2][2][n+1],mod=(long)(1E+9)+7;
                Arrays.fill(dp[1][1],1);dp[1][1][0]=0;
                for(int i=1;i<=k;i++){
                 for(int j=1;j<=n;j++){
                  dp[i][0][j]=(dp[i][0][j-1]+dp[i][1][j-1])%mod;
                  for(int l=j;l<=n;l+=j)dp[i+1][1][l]=(dp[i+1][1][l]+dp[i][1][j])%mod;
                 }
                }
                pw.write((dp[k][0][n]+dp[k][1][n])%mod+"");
		pw.flush();pw.close();
	}
}