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]. 
# name: Stranger001
# date: 15-01-2020
# time: 01:47:05

val = 10**9 + 7
def bin_exp(a, n):
	#gives a**n mod 10**9 +7
	
	if n==0:
		return 1
 
	else:
		
		j = 0
		bin_val = []
		while n >0:
			bin_val.append(n%2)
			n = n//2
			j+=1
		power_arr  =  [a%val]
		for i in range(1,j):
			power_arr.append((power_arr[-1]**2)%val)
 
		value = 1
		for i in range(0, j):
			if bin_val[i] == 1:
				value = (value*power_arr[i])%val
 
		return value%val
		
def fact_arr(n):
	# n th position gives (n)!...i.e 0 th position gives 0!, 1st position gives 1! and so on 
	arr = [1]
	for  i in range(n):
		arr.append((arr[-1]*(i+1))%val)
 
	return arr
	
def mult_inver(a):
	return bin_exp(a, val-2)
 
fact_list = fact_arr(10**3)
inv_fact_list = [1]
for i in range(1, 10**3 +1):
	inv_fact_list.append(mult_inver(fact_list[i]))
 
 
n, m = input().split()
n = int(n)
m = int(m)
val1 = 0
for i in range(1, min(n, 2*m)+1):
	val1= (val1 +(fact_list[n]*inv_fact_list[i]*inv_fact_list[n-i]*fact_list[2*m-1]*inv_fact_list[i-1]*inv_fact_list[2*m-i]))%val
 
print(val1%val)