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 collections import defaultdict
  mo = 998244353
  def solve(n,A):
      d_minus = defaultdict(int)
      d_all = defaultdict(int)
      d_all[-1] = 1
      cnt = 0
      for a in A:
          # if a==0 and d_all[0] == 0:
          #     d_all[0] = 1
          #     cnt += 1
          #     continue
          # if a==1:
          #     if d_all[0] == 0 and d_minus[1]==0:
          #         d_minus[1] = 1
          #         cnt += 1
          #         continue
          cnt += d_all[a]
          d_all[a] = (d_all[a] * 2)%mo
          cnt += d_minus[a]
          d_minus[a] = (d_minus[a]*2)%mo
          cnt += d_all[a-2]
          d_minus[a] = (d_minus[a] + d_all[a-2])%mo
          cnt += d_all[a-1]
          d_all[a] = (d_all[a] + d_all[a-1])%mo
          cnt += d_minus[a+2]
          d_minus[a+2] = (d_minus[a+2]*2)%mo
          # d_all[a] = (d_all[a] + d_minus[a+1])%mo
          cnt = cnt%mo
      return cnt
  
  def main():
      ans = []
      t = int(input())
      for _ in range(t):
          n=int(input())
          A=list(map(int, input().split(' ')))
          ans.append(solve(n,A))
      for a in ans:
          print(a)
  
  
  if __name__ == '__main__':
      main()