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 MAXL = 1 << 22;
  char i_str[MAXL], o_str[MAXL], *i_s, *i_t;
  int o_t;
  inline char gc() {
    if (i_s == i_t) {
      i_s = i_str;
      i_t = i_s + fread(i_str, 1, MAXL, stdin);
      return i_s == i_t ? EOF : *i_s++;
    } else
      return *i_s++;
  }
  inline int read() {
    int x = 0, f = 0;
    char ch = gc();
    for (; ch < '0' || ch > '9'; ch = gc())
      if (ch == '-') f = 1;
    for (; ch >= '0' && ch <= '9'; ch = gc()) x = x * 10 + (ch ^ 48);
    return f ? -x : x;
  }
  inline void pc(char x) {
    o_str[o_t++] = x;
    if (o_t == MAXL) fwrite(o_str, 1, o_t, stdout), o_t = 0;
  }
  void write(int x) {
    if (x < 0) pc('-'), x = -x;
    if (x > 9) write(x / 10);
    pc(x % 10 ^ 48);
  }
  const int N = 5e5 + 5, mod = 998244353;
  int D, f[N][3];
  inline void Solve() {
    int n = read(), ans = 0;
    for (register int i = 0; i <= n + 2; ++i) f[i][0] = f[i][1] = f[i][2] = 0;
    for (register int i = 1; i <= n; ++i) {
      int x = read();
      f[x + 2][1] = (f[x + 2][1] + f[x + 2][1]) % mod;
      if (!x) {
        f[x][0] = (f[x][0] + f[x][0] + 1) % mod;
        continue;
      }
      f[x][0] = ((f[x][0] + f[x][0]) % mod + f[x - 1][0]) % mod;
      if (x == 1) {
        f[x][1] = (f[x][1] + f[x][1] + 1) % mod;
        continue;
      }
      f[x][1] = ((f[x][1] + f[x][1]) % mod + f[x - 2][0]) % mod;
    }
    for (register int i = 0; i <= n + 2; ++i)
      ans = (((ans + f[i][0]) % mod + f[i][1]) % mod + f[i][2]) % mod;
    write(ans), pc('\n');
    return;
  }
  signed main() {
    int T = read();
    while (T--) Solve();
    fwrite(o_str, 1, o_t, stdout), o_t = 0;
    return 0;
  }
  