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 N = 5e5 + 5, mod = 998244353;
  long long f[N][2], a[N], t, n, ans;
  signed main() {
    scanf("%lld", &t);
    while (t--) {
      scanf("%lld", &n);
      for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
      for (long long i = 0; i <= n + 3; i++) f[i][0] = f[i][1] = 0;
      ans = 0;
      for (long long i = 1; i <= n; i++) {
        long long x = f[a[i]][0] + (a[i] == 0);
        if (a[i] - 1 >= 0) x += f[a[i] - 1][0];
        (f[a[i]][0] += x) %= mod;
        long long y = f[a[i] + 1][1];
        (f[a[i] + 1][1] *= 2) %= mod;
        long long z = (a[i] == 1);
        if (a[i] - 2 >= 0) z += f[a[i] - 2][0];
        if (a[i] - 1 >= 0) z += f[a[i] - 1][1], (f[a[i] - 1][1] += z) %= mod;
        (ans += x + y + z) %= mod;
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  