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 = 0x7f7f7f7f;
  const long long linf = 1e18;
  const int maxn = 5e5 + 9;
  const int maxm = 1e5 + 9;
  const double PI = 3.1415926;
  const double eps = 1e-5;
  const int base = 131;
  const int N = 1e6;
  const long long mod = 998244353;
  namespace IO {
  long long read() {
    long long a = 1, b = 0;
    char c = getchar();
    while (c > '9' || c < '0') {
      if (c == '-') a = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      b = (b << 3) + (b << 1) + c - '0';
      c = getchar();
    }
    return a * b;
  }
  void print(long long x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) print(x / 10);
    putchar(x % 10 + '0');
  }
  }  // namespace IO
  using namespace IO;
  int n;
  long long dp[maxn][2];
  int main() {
    int cse = read();
    while (cse--) {
      dp[0][0] = 1;
      n = read();
      for (int i = 1; i <= n; ++i) {
        int x = read() + 1;
        dp[x][0] += dp[x - 1][0] + dp[x][0];
        dp[x][1] += dp[x][1];
        if (x > 1) dp[x][1] += dp[x - 2][0];
        dp[x + 2][1] += dp[x + 2][1];
        dp[x][0] %= mod;
        dp[x][1] %= mod;
        dp[x + 2][1] %= mod;
      }
      long long ans = 0;
      for (int i = 1; i <= n + 1; ++i) {
        ans = (ans + dp[i][0] + dp[i][1]) % mod;
      }
      print(ans);
      putchar('\n');
      for (int i = 0; i <= n + 1; ++i) dp[i][0] = dp[i][1] = 0;
    }
    return 0;
  }
  