description: Given n, find any array a_1, a_2, …, a_n of integers such that all of the following conditions hold: 
  
    * 1 ≤ a_i ≤ 10^9 for every i from 1 to n.
  
    * a_1 < a_2 < … <a_n
  
    * For every i from 2 to n, a_i isn't divisible by a_{i-1}
  
  
  
  
  It can be shown that such an array always exists under the constraints of the problem.
  
  Input
  
  The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
  
  The only line of each test case contains a single integer n (1 ≤ n ≤ 1000).
  
  It is guaranteed that the sum of n over all test cases does not exceed 10^4.
  
  Output
  
  For each test case print n integers a_1, a_2, …, a_n — the array you found. If there are multiple arrays satisfying all the conditions, print any of them.
  
  Example
  
  Input
  
  
  3
  1
  2
  7
  
  
  Output
  
  
  1
  2 3
  111 1111 11111 111111 1111111 11111111 111111111
  
  Note
  
  In the first test case, array [1] satisfies all the conditions.
  
  In the second test case, array [2, 3] satisfies all the conditions, as 2<3 and 3 is not divisible by 2.
  
  In the third test case, array [111, 1111, 11111, 111111, 1111111, 11111111, 111111111] satisfies all the conditions, as it's increasing and a_i isn't divisible by a_{i-1} for any i from 2 to 7.
solution: import collections
  import sys
  from collections import OrderedDict,Counter, defaultdict
  import  math
  import bisect
  from sys import stdin, stdout
  
  def with_index():
      a = []
      i = 0
      for t in map(int, input().split()):
          a.append((t,i))
          i+= 1
      return a
  def get_line(): return list(map(int, sys.stdin.readline().strip().split()))
  def in1(): return int(input())
  def take_input():
      return stdin.readline()
  def display_arr(arr):
      stdout.write(" ".join(map(str, arr)) + "\n")
  def display_num(num):
      stdout.write(str(num) + "\n")
  def factorial(n): return n * factorial(n-1) if n>1 else 1
  def percent(c,s): return (c/s) * 100
  def l2d(a):
      d= {}
      for i in a:
          if i in d: d[i] += 1
          else: d[i] = 1
      return d
  
  def min_op(a, n, count):
      if n== 1: return 0
      if count >= 2:
          return n - count//2
      return -1
  def Arr(n):
      if n == 1: return [1]
      if n == 2: return [2,3]
      a = [2,3]
      i = 4
      while len(a) !=n:
          if i % a[-1] !=0:
              a.append(i)
          i += 1
      return a
  
  for _ in range(in1()):
      n = in1()
      print(*Arr(n))