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 int inf = 0x3f3f3f3f;
  long long fpow(long long n, long long k, long long m = 998244353) {
    if (k == 1) return n % m;
    if (k == 0) return 1 % m;
    long long t = fpow(n % m, k / 2, m);
    return (k % 2 == 0) ? ((t * t) % m) : (((t * t % m) * (n % m)) % m);
  }
  int q;
  long long n, a[500005], cnt1[500005], cnt2[500005], c1;
  void exeCute() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    memset(cnt1, 0, sizeof(long long) * (n + 2));
    memset(cnt2, 0, sizeof(long long) * (n + 2));
    c1 = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] == 0)
        cnt1[a[i]] = (cnt1[a[i]] * 2 + 1) % 998244353;
      else {
        cnt1[a[i]] = (cnt1[a[i]] * 2 + cnt1[a[i] - 1]) % 998244353;
      }
      cnt2[a[i]] = (cnt2[a[i]] * 2) % 998244353;
      if (a[i] >= 2) cnt2[a[i]] = (cnt2[a[i]] + cnt1[a[i] - 2]) % 998244353;
      if (cnt2[a[i] + 2] > 0) cnt2[a[i] + 2] = (cnt2[a[i] + 2] * 2) % 998244353;
      c1 += (a[i] == 1);
    }
    long long ans = (fpow(2, c1) - 1 + 998244353) % 998244353;
    for (int i = 0; i <= n; i++) ans = (ans + cnt1[i] + cnt2[i]) % 998244353;
    cout << ans << endl;
  }
  void init() {
    cin >> q;
    for (int i = 0; i < q; i++) exeCute();
  }
  int main(int argc, char* argv[]) {
    init();
    return 0;
  }
  