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;
  const long long int mod = 998244353;
  const long long int nmax = 2 * 1e3 + 10;
  vector<pair<long long int, long long int> > dir{
      {0, -1}, {-1, 0}, {0, 1}, {1, 0}};
  long long int mod_sub(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  long long int mod_add(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long int mod_mul(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  void sieve(vector<long long int>& arr, long long int n) {
    arr[0] = -1;
    arr[1] = -1;
    for (long long int i = 2; (i * i) <= n; i++) {
      if (arr[i] != -1) {
        for (long long int j = i * i; j <= n; j += i) arr[j] = -1;
      }
    }
  }
  long long int gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long int mod_exp(long long int a, long long int b, long long int c) {
    if (a == 0) return 0;
    if (b == 0) return 1;
    long long int smallans;
    if (b % 2 == 0) {
      smallans = mod_exp(a, b / 2, c);
      return mod_mul(smallans, smallans, c);
    } else {
      smallans = mod_exp(a, b - 1, c);
      return mod_mul(a, smallans, c);
    }
  }
  void solve() {
    long long int n;
    cin >> n;
    vector<long long int> a(n + 1);
    for (long long int i = 1; i < n + 1; i++) {
      cin >> a[i];
    }
    vector<long long int> dp1(n + 2);
    dp1[0] = 1;
    vector<long long int> dp2(n + 2);
    for (long long int i = 1; i < n + 1; i++) {
      if (a[i] > 0) {
        dp2[a[i] - 1] = (dp1[a[i] - 1] % mod + (2 * dp2[a[i] - 1]) % mod) % mod;
      }
      dp1[a[i] + 1] = (dp1[a[i]] % mod + (2 * dp1[a[i] + 1]) % mod) % mod;
      dp2[a[i] + 1] = (2 * dp2[a[i] + 1]) % mod;
    }
    long long int sum = 0;
    for (long long int i = 0; i < n + 1; i++) {
      sum = (sum % mod + dp1[i] % mod + dp2[i] % mod) % mod;
    }
    cout << ((sum - 1) % mod + mod) % mod;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long int t = 1;
    cin >> t;
    while (t--) {
      solve();
      cout << "\n";
    }
  }
  