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 Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        String s[]=bu.readLine().split(" ");
        int n=Integer.parseInt(s[0]),m=Integer.parseInt(s[1]),i;
        long f[]=new long[2001],M=1000000007;
        f[0]=1;
        for(i=1;i<=2000;i++)
        f[i]=f[i-1]*i%M;
        //ans=(n+2m-1)!/(n-1)!(2m)!

        long den=f[2*m]*f[n-1]%M,num=f[n+2*m-1];
        System.out.print(num*power(den,M-2,M)%M);
    }

    static long power(long a,long b,long M)
    {
        long res=1;
        while(b!=0)
        {
            if(b%2==1) res=res*a%M;
            b>>=1;
            a=a*a%M;
        }
        return res;
    }
}
