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]. 
n, mm = map(int,input().split())
mod = 10**9+7

l = 10**4

M = [1]  # i!のmod
m = 1
for i in range(1, l):
    m = (m * i) % mod
    M.append(m)

def pow(x, y, mod):  # x**y の mod を返す関数
    ans = 1
    while y > 0:
        if y % 2 == 1:
            ans = (ans * x) % mod
        x = (x**2) % mod
        y //= 2
    return ans

def inv(x, mod):  # x の mod での逆元を返す関数
    return pow(x, mod-2, mod)

# print(inv(8,13))

MI = [0] * (l-1) +[inv(M[l-1], mod)]  # i!の逆元
for i in range(l-2, -1, -1):
    MI[i] = MI[i+1] * (i+1) % mod

def C(x, y):  # コンビネーション
    if y < 0 or y > x:
        return 0
    ans = M[x]
    ans = (ans * MI[y]) % mod
    return (ans * MI[x-y]) % mod

ans = 0
for i in range(n):
    # print((mm-1+i, i),(mm, n-i-1))
    ans = (ans + C(mm-1+i, i) * C(mm+n-i-1, n-i-1)) % mod
print(ans)