description: Let's call a sequence of integers x_1, x_2, ..., x_k MEX-correct if for all i (1 ≤ i ≤ k) |x_i - \operatorname{MEX}(x_1, x_2, ..., x_i)| ≤ 1 holds. Where \operatorname{MEX}(x_1, ..., x_k) is the minimum non-negative integer that doesn't belong to the set x_1, ..., x_k. For example, \operatorname{MEX}(1, 0, 1, 3) = 2 and \operatorname{MEX}(2, 1, 5) = 0.
  
  You are given an array a consisting of n non-negative integers. Calculate the number of non-empty MEX-correct subsequences of a given array. The number of subsequences can be very large, so print it modulo 998244353. 
  
  Note: a subsequence of an array a is a sequence [a_{i_1}, a_{i_2}, ..., a_{i_m}] meeting the constraints 1 ≤ i_1 < i_2 < ... < i_m ≤ n. If two different ways to choose the sequence of indices [i_1, i_2, ..., i_m] yield the same subsequence, the resulting subsequence should be counted twice (i. e. two subsequences are different if their sequences of indices [i_1, i_2, ..., i_m] are not the same).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 5 ⋅ 10^5).
  
  The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ n).
  
  The sum of n over all test cases doesn't exceed 5 ⋅ 10^5.
  
  Output
  
  For each test case, print a single integer — the number of non-empty MEX-correct subsequences of a given array, taken modulo 998244353.
  
  Example
  
  Input
  
  
  4
  3
  0 2 1
  2
  1 0
  5
  0 0 0 0 0
  4
  0 1 2 3
  
  
  Output
  
  
  4
  2
  31
  7
  
  Note
  
  In the first example, the valid subsequences are [0], [1], [0,1] and [0,2].
  
  In the second example, the valid subsequences are [0] and [1].
  
  In the third example, any non-empty subsequence is valid. 
solution: from sys import stdin, stdout
  
  MOD = 998244353
  
  
  def solve_test(n, elements):
      dp_increasing = [0] * (n + 1)
      dp_alternate = [0] * (n + 1)
      # dp_increasing[i] = # subsequences that end on element i
      # dp_alternate[i] = # subsequences that alternate with mex value equal to i (mex can only be 0 1 2 3 .. n - 1
      # 0 1 2 3 4 .. x-2 x
  
      for x in elements:
          dp_increasing[x] = (dp_increasing[x] * 2) % MOD  # 0 1 2 .. x | add x
          if x >= 1:
              dp_increasing[x] = (dp_increasing[x] + dp_increasing[x - 1]) % MOD  # 0 1 2 3 .. x-1 | add x
          if x == 0:
              dp_increasing[0] = (dp_increasing[0] + 1) % MOD  # we have subsequence of length 1 for [0]
  
          # add x to a sequence alternating with MEX (x-1) or (x+1)
          if x >= 1:
              dp_alternate[x - 1] = (dp_alternate[x - 1] * 2) % MOD
          if x + 1 < n + 1:
              dp_alternate[x + 1] = (dp_alternate[x + 1] * 2) % MOD
  
          # 0 1 2 .. x - 2 | add x ( form another sequence) that alternates with MEX (x-1)
          if x >= 2:
              dp_alternate[x - 1] = (dp_alternate[x - 1] + dp_increasing[x - 2]) % MOD
  
          if x == 1:
              dp_alternate[0] = (dp_alternate[0] + 1) % MOD  # we can have [1]
      answer = [(x + y) % MOD for x, y in zip(dp_increasing, dp_alternate)]
      return sum(answer) % MOD
  
  
  def stress_test():
      n = int(1e5)
      elements = [i for i in range(n)]
      solve_test(n, elements)
  
  
  def read_input():
      tests = int(stdin.readline())
      for test in range(tests):
          n = int(stdin.readline())
          elements = [int(x) for x in stdin.readline().split()]
          stdout.write(str(solve_test(n, elements)) + '\n')
  
  
  def read_file():
      with open("input.txt") as f:
          tests = int(f.readline())
          for test in range(tests):
              n = int(f.readline())
              elements = [int(x) for x in f.readline().split()]
              print(solve_test(n, elements))
  
  
  # Press the green button in the gutter to run the script.
  if __name__ == '__main__':
      read_input()
  