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: import sys
  
  # sys.setrecursionlimit(200005)
  int1 = lambda x: int(x)-1
  p2D = lambda x: print(*x, sep="\n")
  def II(): return int(sys.stdin.readline())
  def LI(): return list(map(int, sys.stdin.readline().split()))
  def LLI(rows_number): return [LI() for _ in range(rows_number)]
  def LI1(): return list(map(int1, sys.stdin.readline().split()))
  def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
  def SI(): return sys.stdin.readline().rstrip()
  dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
  # dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
  # inf = 18446744073709551615
  inf = 4294967295
  # md = 10**9+7
  md = 998244353
  
  def solve(n, aa):
      m1 = [0]*(n+2)
      m2 = [0]*(n+2)
      m1[0] = 1
      ans = 0
      for a in aa:
          if a-1 >= 0:
              ans += m1[a-1]+m2[a-1]
              m2[a-1] += m1[a-1]+m2[a-1]
              m2[a-1] %= md
          ans += m1[a]+m1[a+1]+m2[a+1]
          m1[a+1] += m1[a]+m1[a+1]
          m2[a+1] *= 2
          m1[a+1] %= md
          m2[a+1] %= md
          ans %= md
          # print(a, ans)
          # print(m1)
          # print(m2)
      print(ans)
  
  def solve2(n, aa):
      def ok(bit):
          st = set()
          for i, a in enumerate(aa):
              if bit >> i & 1 == 0: continue
              st.add(a)
              for b in range(n+1):
                  if b not in st:
                      if abs(b-a) > 1: return False
                      break
          return True
  
      ans = 0
      for bit in range(1, 1 << n):
          if ok(bit):
              ans += 1
              print(bin(bit)[2:].zfill(n)[::-1])
      print(ans)
  
  for testcase in range(II()):
      n = II()
      aa = LI()
      solve(n, aa)
      # solve2(n,aa)
  