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: 
  def solve(n, seq):
      mod = 998244353
      dynup = [0]*(n+3)
      dyndown = [0]*(n+3)
  
      # dyn[0] = 1
      for i,x in enumerate(seq):
          if x == 0:
              # dyndown[2] = (dyndown[2] + dynup[0])%mod
              dyndown[1] = (2*dyndown[1] + 1)%mod
              dynup[1] = (dynup[1] + dynup[1])%mod
          elif x == 1:
              # dyndown[3] = (dyndown[3] + dynup[1])%mod
              dyndown[2] = (dyndown[2] + dyndown[2] + dyndown[1])%mod
              dynup[2] = (dynup[2] + dynup[2])%mod
              dynup[0] = (dynup[0] + dynup[0] + 1)%mod
  
          else:
              # dyndown[x+2] = (dynup[x] + dyndown[x+2])%mod
              dyndown[x+1] = (dyndown[x+1] + dyndown[x+1]+ dyndown[x])%mod
              # dyndown[x-1] if we add x and mex is x-1 then this is a dynup
              dynup[x-1] = (dynup[x-1] + dynup[x-1] + dyndown[x-1])%mod
              dynup[x+1] = (dynup[x+1] + dynup[x+1])%mod
      ans = 0
      for x in dynup:
          ans = (ans +x )%mod
      for x in dyndown:
          ans = (ans +x )%mod
  
      return ans
  
  
  import os
  import io
  # import time
  # a=time.time()
  if __name__ == "__main__":
      input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  
  
      T = int(input().decode().strip())
      for t in range(T):
          n = int(input().decode().strip())
          seq = [int(x) for x in input().decode().strip().split(" ")]
          assert len(seq) == n
          res=solve(n, seq)
          print(res, flush=True)
  