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 math
  import sys
  import operator as op
  from functools import reduce
  from collections import OrderedDict
  
  
  class Input:
      @staticmethod
      def string():
          return str(input())
  
      @staticmethod
      def int():
          return int(input())
  
      @staticmethod
      def list(sep=' '):
          return input().split(sep)
  
      @staticmethod
      def list_int(sep=' '):
          return list(map(int, input().split(sep)))
  
  
  class Output:
      @staticmethod
      def print(res):
          print(res)
  
      @staticmethod
      def string(res):
          print(str(res))
  
      @staticmethod
      def list(res):
          print(' '.join(str(x) for x in res))
  
  
  class Algorithms:
      @staticmethod
      def isPrime(n):
          return all(n % i != 0 for i in range(2, int(math.sqrt(n) + 1)))
  
      @staticmethod
      def getPrimeNumbers(n):
          a = [1]
          for num in range(2, n):
              if all(num % i != 0 for i in range(2, int(math.sqrt(num) + 1))):
                  a.append(num)
          return a
  
      # Факторизация - Разложение числа на простые множители
      @staticmethod
      def factorization(n):
          res = {}
          for i in range(2, n * n):
              cnt = 0
              while n % i == 0:
                  cnt += 1
                  n = n / i
              if cnt > 0:
                  res[i] = cnt
          if n > 1:
              res[n] = 1
          return res
  
      # Наибольший общий делитель (НОД)
      @staticmethod
      def gcd(a, b):
          if b == 0:
              return a
          else:
              a = a % b
              return Algorithms.gcd(b, a)
  
      # Наименьшее общее кратное (НОК)
      @staticmethod
      def lcm(a, b):
          return abs(a * b) // Algorithms.gcd(a, b)
  
      # Последовательность Фибоначчи =  0, 1, 1, 2, 3, 5, 8, 13, 21, 34
      @staticmethod
      def fibonacci(n):
          a = [0, 1]
          for i in range(2, n + 1):
              a.append(a[i - 1] - a[i - 2])
          return a[n]
  
      @staticmethod
      def z_func(s):
          n = len(s)
          z = [0] * n
          l = r = 0
          for i in range(1, n):
              if r >= i:
                  z[i] = min(z[i - l], r - i + 1)
              while z[i] + i < n and s[z[i]] == s[z[i] + i]:
                  z[i] += 1
              if i + z[i] - 1 > r:
                  l = i
                  r = i + z[i] - 1
          return z
  
      @staticmethod
      def faktorial(n):
          factorial = 1
          for i in range(3, n + 1):
              factorial *= i
              factorial = factorial % (math.pow(10, 9) + 7)
  
          return factorial
  
  def solve():
      n = Input.int()
      p = 2 * n
      print(int(Algorithms.faktorial(p)))
  
  
  #query_count = 1
  query_count = Input.int()
  while query_count:
      query_count -= 1
      solve()