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 final class Today {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt(),m=sc.nextInt();
		o(nCrModpLucas(n+2*m-1,n-1,1000000007));
	}

	static void o(Object s) {
		System.out.println(s);
	}

	static int nCrModpDP(int n, int r, int p) 
	{ 
		int[] C=new int[r+1]; 

		C[0] = 1;
		for (int i = 1; i <= n; i++) 
		{ 
			for (int j = Math.min(i, r); j > 0; j--) 
				C[j] = (C[j] + C[j-1])%p; 
		} 
		return C[r]; 
	} 
	static int nCrModpLucas(int n, int r, int p) 
	{ 
	if (r==0) 
		return 1; 
	int ni = n%p; 
	int ri = r%p; 
	return (nCrModpLucas(n/p, r/p, p) * // Last digits of n and r 
			nCrModpDP(ni, ri, p)) % p; // Remaining digits 
	}
}
