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 gcd(long long a, long long b) { return a % b ? gcd(b, a % b) : b; }
  const long long mod = 998244353;
  const long long inf = 1e18;
  long long qpow(long long a, long long b) {
    long long t = 1;
    while (b != 0) {
      if (b & 1) t = (t * a) % mod;
      a = (a * a) % mod;
      b >>= 1;
    }
    return t;
  }
  template <typename T>
  void re(T& a) {
    for (auto& t : a) cin >> t;
  }
  template <typename T>
  void pr(T& a) {
    for (auto& t : a) cout << t << " ";
    cout << endl;
  }
  const int N = 5e5 + 5;
  long long a[N];
  long long dp[N];
  long long pd[N];
  int n;
  void add(long long& x, long long y) { x = ((x + y) % mod + mod) % mod; }
  void solve() {
    cin >> n;
    for (int i = 0; i <= n; ++i) {
      dp[i] = 0;
      pd[i] = 0;
    }
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
    }
    for (int i = 1; i <= n; ++i) {
      long long x = a[i];
      long long tmp = dp[x];
      if (x == 0) {
        add(tmp, 1);
      } else {
        add(tmp, dp[x - 1]);
      }
      add(dp[x], tmp);
      x += 2;
      add(pd[x], pd[x]);
      x -= 2;
      if (x == 0) {
        continue;
      }
      tmp = pd[x];
      if (x == 1) {
        add(tmp, 1);
      } else {
        add(tmp, dp[x - 2]);
      }
      add(pd[x], tmp);
    }
    long long ans = 0;
    for (int i = 0; i <= n; ++i) {
      add(ans, dp[i]);
      add(ans, pd[i]);
    }
    cout << ans << endl;
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  