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;
  long long max(long long a, long long b) {
    if (a > b) {
      return a;
    } else {
      return b;
    }
  }
  long long min(long long a, long long b) {
    if (a < b) {
      return a;
    } else {
      return b;
    }
  }
  long long rand_int(long long l, long long r) {
    static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
    return uniform_int_distribution<long long>(l, r)(gen);
  }
  long long power(long long b, long long e) {
    if (e == 0) return 1;
    if (e % 2)
      return ((b * power((b) * (b), (e - 1) / 2)));
    else
      return power((b) * (b), e / 2);
  }
  long long modpower(long long b, long long e, long long q) {
    long long MOD = q;
    if (e == 0) return 1;
    if (e % 2)
      return ((b % MOD) * modpower((b % MOD) * (b % MOD), (e - 1) / 2, q)) % MOD;
    else
      return modpower((b % MOD) * (b % MOD), e / 2, q) % MOD;
  }
  void dpv(vector<long long> v) {
    for (long long i = 0; i < v.size(); i++) {
      cout << v[i] << " ";
    }
    cout << '\n';
  }
  void dpv(vector<pair<long long, long long> > v) {
    for (long long i = 0; i < v.size(); i++) {
      cout << v[i].first << " " << v[i].second << '\n';
    }
  }
  void dpv(set<long long> v) {
    for (auto i : v) {
      cout << i << " ";
    }
    cout << '\n';
  }
  long long ceil1(long long x, long long y) {
    long long r = x / y;
    if (x % y == 0) {
      return r;
    } else {
      return r + 1;
    }
  }
  const long long md = 998244353;
  void add(long long &x, long long y) {
    x += y;
    if (x >= md) {
      x -= md;
    }
  }
  void oblivious() {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<long long> dp1(n + 5);
    vector<long long> dp2(n + 5);
    dp1[0] = 1;
    for (long long i = 0; i < n; i++) {
      long long x = a[i];
      add(dp1[x + 1], dp1[x + 1]);
      add(dp1[x + 1], dp1[x]);
      if (x > 0) {
        add(dp2[x - 1], dp2[x - 1]);
      }
      if (x > 0) {
        add(dp2[x - 1], dp1[x - 1]);
      }
      add(dp2[x + 1], dp2[x + 1]);
    }
    long long ans = 0;
    for (auto i : dp2) {
      add(ans, i);
    }
    for (auto i : dp1) {
      add(ans, i);
    }
    cout << (ans - 1 + md) % md << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    while (t--) {
      oblivious();
    }
    return 0;
  }
  