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]. 
#!/usr/bin/env python3
import os
import sys
from atexit import register
from io import StringIO

# Buffered output, at least 2x faster.
sys.stdout = StringIO()
register(lambda: os.write(1, sys.stdout.getvalue().encode()))

###############################################################################
import math
###############################################################################
MOD = int(1e9 + 7)

n,m = [int(k) for k in input().split()]

dp = {}
def solve(size, cap):
    key = (size, cap)
    if key in dp:
        return dp[key]
    if size == 0:
        dp[key] = 1
        return dp[key]
    elif size == 1:
        dp[key] = cap
        return dp[key]
    else:
        ret = 0
        for i in range(1, cap+1):
            ret = (ret + solve(size-1, i))%MOD
        dp[key] = ret
        return dp[key]

for i in range(m):
    for j in range(n):
        solve(m,n)

ret = 0
for i in range(1, n+1):
    for j in range(i, n+1):
        k = (solve(m-1, i) * solve(m-1, n+1-j)) % MOD
        ret = (ret + k) % MOD

print(ret)
