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.util.*;
import java.io.*;

public class problem1288 {

	static StreamTokenizer in;

	static int nextInt() throws Exception {
		in.nextToken();
		return (int) in.nval;
	}

	static String next() throws Exception {
		in.nextToken();
		return (String) in.sval;
	}

	public static void main(String[] args) throws Exception {
	//	in = new StreamTokenizer(new BufferedReader(new FileReader("test.in")));
		    in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		int n=nextInt(), m=nextInt();
		long [][]dp=new long[n+1][m+1];
		long mod=1_000_000_000+7;
		// value length
		// dp=max
		for(int i=1;i<=n;i++) {
			dp[i][1]=1;
		}
		for(int j=1;j<=m;j++) {
			dp[1][j]=1;
		}
		for(int i=2;i<=n;i++) {
			for(int j=2;j<=m;j++) {
				for(int k=1;k<=i-1;k++) {
					dp[i][j]+=dp[k][j-1];
					dp[i][j]%=mod;
				}
				dp[i][j]+=dp[i][j-1];
				dp[i][j]%=mod;
			}
		}
		long ret=0;
		for(int i=n;i>=1;i--) {
			int min=i;
			long sum=0;
			for(int j=1;j<=min;j++) {
				int max=j;
				sum+=dp[max][m];
				sum%=mod;
			}
			ret+=sum*dp[n-min+1][m];
			ret%=mod;
		}
		System.out.println(ret);
	}
}

