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: #include <bits/stdc++.h>
  using namespace std;
  using LL = long long;
  int cases = -1;
  enum InputType : int {
    SingleInput = 1,
    MultipleInput = 2,
    NotSet = 0,
  };
  void getCases() {
    InputType inputType = MultipleInput;
    if (inputType == SingleInput) {
      cases = 1;
    } else if (inputType == MultipleInput) {
      scanf("%d", &cases);
    } else {
      assert(false && "fix the function");
    }
  }
  int n;
  int ar[500005];
  LL dp[2][500005];
  const LL MOD = 998244353;
  void Input() {
    scanf("%d", &n);
    for (int i = 0; i <= n + 3; ++i) {
      dp[0][i] = dp[1][i] = 0;
    }
    for (int i = 0; i < n; ++i) {
      scanf("%d", &ar[i]);
    }
  }
  LL add(LL x, LL y) { return (x + y) % MOD; }
  int main() {
    getCases();
    while (cases--) {
      Input();
      dp[0][0] = 1;
      for (int i = 0; i < n; ++i) {
        int x = ar[i];
        dp[0][x + 1] = add(dp[0][x + 1], dp[0][x + 1]);
        dp[0][x + 1] = add(dp[0][x + 1], dp[0][x]);
        if (x > 0) {
          dp[1][x - 1] = add(dp[1][x - 1], dp[1][x - 1]);
          dp[1][x - 1] = add(dp[1][x - 1], dp[0][x - 1]);
        }
        dp[1][x + 1] = add(dp[1][x + 1], dp[1][x + 1]);
      }
      LL ans = 0;
      for (int i = 0; i <= n + 3; ++i) {
        ans = add(ans, dp[0][i]);
        ans = add(ans, dp[1][i]);
      }
      ans = add(ans, MOD - 1);
      printf("%lld\n", ans);
    }
  }
  