You are given two integers n and m. Calculate the number of pairs of arrays (a, b) such that:

  * the length of both arrays is equal to m; 
  * each element of each array is an integer between 1 and n (inclusive); 
  * a_i ≤ b_i for any index i from 1 to m; 
  * array a is sorted in non-descending order; 
  * array b is sorted in non-ascending order. 



As the result can be very large, you should print it modulo 10^9+7.

Input

The only line contains two integers n and m (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10).

Output

Print one integer – the number of arrays a and b satisfying the conditions described above modulo 10^9+7.

Examples

Input


2 2


Output


5


Input


10 1


Output


55


Input


723 9


Output


157557417

Note

In the first test there are 5 suitable arrays: 

  * a = [1, 1], b = [2, 2]; 
  * a = [1, 2], b = [2, 2]; 
  * a = [2, 2], b = [2, 2]; 
  * a = [1, 1], b = [2, 1]; 
  * a = [1, 1], b = [1, 1]. 
import java.io.*;
import java.util.*;

public class  Main{
	static long[][]memo1,memo2;
	static long mod=(long)1e9+7;
	static int max;
	static long dp(int i,int prev) {
		if(i<0)return 1;
		if(memo1[i][prev]!=-1)return memo1[i][prev];
		long ans=0;
		for(int j=prev;j<=max;j++) {
			ans=(ans+dp(i-1,j))%mod;
		}
		return memo1[i][prev]=ans;
	}
	static long dp2(int i,int prev) {
		if(i<0)return 1;
		if(memo2[i][prev]!=-1)return memo2[i][prev];
		long ans=0;
		for(int j=prev;j>=1;j--) {
			ans=(ans+dp2(i-1,j))%mod;
		}
		return memo2[i][prev]=ans;
	}
	public static void main(String[] args) throws Exception{
		MScanner sc=new MScanner(System.in);
		PrintWriter pw=new PrintWriter(System.out);
		int n=sc.nextInt(),m=sc.nextInt();
		max=n;	
		memo1=new long[m][n+1];
		for(long[]i:memo1) {
			Arrays.fill(i,-1);
		}
		memo2=new long[m][n+1];
		for(long[]i:memo2) {
			Arrays.fill(i,-1);
		}
		long ans=0;
		for(int i=1;i<=n;i++) {
			long arraysB=dp(m-2,i);
			for(int j=i;j>=1;j--) {
				long arraysA=dp2(m-2,j);
				
				long inc=(arraysA*arraysB)%mod;
				ans=(ans+inc)%mod;
			}
		}
		pw.println(ans);
		pw.flush();
	}	
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int[] takearr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] takearrl(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public Integer[] takearrobj(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] takearrlobj(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}