description: CQXYM is counting permutations length of 2n.
  
  A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
  
  A permutation p(length of 2n) will be counted only if the number of i satisfying p_i<p_{i+1} is no less than n. For example:
  
    * Permutation [1, 2, 3, 4] will count, because the number of such i that p_i<p_{i+1} equals 3 (i = 1, i = 2, i = 3).
    * Permutation [3, 2, 1, 4] won't count, because the number of such i that p_i<p_{i+1} equals 1 (i = 3). 
  
  
  
  CQXYM wants you to help him to count the number of such permutations modulo 1000000007 (10^9+7).
  
  In addition, [modulo operation](https://en.wikipedia.org/wiki/Modulo_operation) is to get the remainder. For example:
  
    * 7 mod 3=1, because 7 = 3 ⋅ 2 + 1, 
    * 15 mod 4=3, because 15 = 4 ⋅ 3 + 3. 
  
  Input
  
  The input consists of multiple test cases. 
  
  The first line contains an integer t (t ≥ 1) — the number of test cases. The description of the test cases follows.
  
  Only one line of each test case contains an integer n(1 ≤ n ≤ 10^5).
  
  It is guaranteed that the sum of n over all test cases does not exceed 10^5
  
  Output
  
  For each test case, print the answer in a single line.
  
  Example
  
  Input
  
  
  4
  1
  2
  9
  91234
  
  
  Output
  
  
  1
  12
  830455698
  890287984
  
  Note
  
  n=1, there is only one permutation that satisfies the condition: [1,2].
  
  In permutation [1,2], p_1<p_2, and there is one i=1 satisfy the condition. Since 1 ≥ n, this permutation should be counted. In permutation [2,1], p_1>p_2. Because 0<n, this permutation should not be counted.
  
  n=2, there are 12 permutations: [1,2,3,4],[1,2,4,3],[1,3,2,4],[1,3,4,2],[1,4,2,3],[2,1,3,4],[2,3,1,4],[2,3,4,1],[2,4,1,3],[3,1,2,4],[3,4,1,2],[4,1,2,3].
solution: # import collections
  # import random
  # import math
  import itertools
  # import math
  # mport math
  from collections import defaultdict
  import itertools
  # from sys import stdin, stdout
  import math
  import sys
  
  import bisect
  # import operator
  # from decimal import Decimal
  
  # sys.setrecursionlimit(10**6)
  
  p2D = lambda x: print(*x, sep="\n")
  def II(): return int(sys.stdin.buffer.readline())
  def MI(): return map(int, sys.stdin.buffer.readline().split())
  def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
  def LLI(rows_number): return [LI() for _ in range(rows_number)]
  def BI(): return sys.stdin.buffer.readline().rstrip()
  def SI(): return sys.stdin.buffer.readline().rstrip().decode()
  def li(): return [int(i) for i in input().split()]
  def lli(rows): return [li() for _ in range(rows)]
  def si(): return input()
  def ii(): return int(input())
  def ins(): return input().split()
  
  
  
  def solve():
      n = II()
      mod = 10**9+7
      ans = 1
      for i in range(3,2*n+1):
          ans= (ans*i)%mod
      return ans
  
  
  
  
  def main():
      #print('finish '+ str(solve()), flush=True)
      for _ in range(II()):
          sys.stdout.write(str(solve()) + "\n")
      #sys.exit(0)
      #print(solve())
  
      #     z += str(ans) + '\n'
      # print(len(ans), ' '.join(map(str, ans)), sep='\n')
      # stdout.write(z)
  
  
  # for interactive problems
  # print("? {} {}".format(l,m), flush=True)
  # or print this after each print statement
  # sys.stdout.flush()
  
  
  if __name__ == "__main__":
      main()