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.lang.*;
import java.io.*;

public class Main
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
  
    void solve() throws IOException {
        int n=ni(),m=ni();
        long mod=1000000007;
        long[][]A=new long[m+1][n+1];
        for (int x=1;x<=n;x++) A[m][x]=1;
        long prevsum=n;
        for (int x=m-1;x>0;x--) {
            long newsum=0;
            for (int y=1;y<=n;y++) {
                A[x][y]=prevsum;
                newsum=(newsum+A[x][y])%mod;
                prevsum=(prevsum-A[x+1][y]+mod)%mod;
            }
            prevsum=newsum;
        }
        
        long ans=A[1][1];
        prevsum=1;
        for (int x=2;x<=n;x++) {
            prevsum=(prevsum+A[1][n-x+1])%mod;
            ans=(ans+(A[1][x]*prevsum)%mod)%mod;
        }
        System.out.println(ans);
    }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}