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 MAXN = 1e6 + 10;
  const long long INF = 1e9 + 10;
  long long MOD = 998244353;
  long double EPS = 1e-12;
  void fastIO() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  long long t, n, a[MAXN];
  long long dp1[MAXN], dp2[MAXN], cnt[MAXN];
  int sum(int x, int y) {
    x += y;
    if (x >= MOD) x -= MOD;
    return x;
  }
  int mul(int x, int y) { return x * 1ll * y % MOD; }
  int bin_pow(int x, int n) {
    int res = 1;
    while (n > 0) {
      if (n & 1) res = mul(res, x);
      x = mul(x, x);
      n >>= 1;
    }
    return res;
  }
  int main() {
    fastIO();
    cin >> t;
    while (t--) {
      cin >> n;
      fill(cnt, cnt + n + 1, 0);
      for (int i = 0; i < n; i++) {
        cin >> a[i];
        cnt[a[i]]++;
      }
      fill(dp1, dp1 + n + 10, 0);
      fill(dp2, dp2 + n + 10, 0);
      int ans = 0;
      for (int i = 0; i < n; i++) {
        int val = a[i];
        cnt[val]--;
        dp1[val] = sum(dp1[val], dp1[val]);
        if (val > 0) {
          dp1[val] = sum(dp1[val], dp1[val - 1]);
        } else {
          dp1[val] = sum(dp1[val], 1);
        }
        if (val > 1) {
          ans = sum(ans, mul(dp1[val - 2], bin_pow(2, cnt[val - 2] + cnt[val])));
        }
        if (val == 1) {
          ans = sum(ans, bin_pow(2, cnt[val]));
        }
      }
      for (int i = 0; i <= n; i++) {
        ans = sum(ans, dp1[i]);
      }
      cout << ans << '\n';
    }
    return 0;
  }
  