#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long myRand(long long B) { return (unsigned long long)rng() % B; }
void swap(long long& x, long long& y) { (x ^= y), (y ^= x), (x ^= y); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    long long dp[n + 2][2];
    for (int i = 0; i < n + 2; i++)
      for (int j = 0; j < 2; j++) dp[i][j] = 0;
    long long inp;
    dp[0][0] = 1;
    for (int i = 0; i < n; i++) {
      cin >> inp;
      dp[inp + 1][0] =
          (dp[inp][0] + 2 * dp[inp + 1][0] % 998244353) % 998244353;
      dp[inp + 1][1] = (dp[inp + 1][1] * 2) % 998244353;
      if (inp > 0)
        dp[inp - 1][1] =
            (dp[inp - 1][0] + dp[inp - 1][1] * 2 % 998244353) % 998244353;
    }
    long long ans = 0;
    for (int i = 0; i < n + 1; i++) {
      ans = (ans +
             ((i != 0 ? dp[i][0] : 0) + (i != n ? dp[i][1] : 0)) % 998244353) %
            998244353;
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long myRand(long long B) { return (unsigned long long)rng() % B; }
void swap(long long& x, long long& y) { (x ^= y), (y ^= x), (x ^= y); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    long long dp[n + 2][2];
    for (int i = 0; i < n + 2; i++)
      for (int j = 0; j < 2; j++) dp[i][j] = 0;
    long long inp;
    dp[0][0] = 1;
    for (int i = 0; i < n; i++) {
      cin >> inp;
      dp[inp + 1][0] =
          (dp[inp][0] + 2 * dp[inp + 1][0] % 998244353) % 998244353;
      dp[inp + 1][1] = (dp[inp + 1][1] * 2) % 998244353;
      if (inp > 0)
        dp[inp - 1][1] =
            (dp[inp - 1][0] + dp[inp - 1][1] * 2 % 998244353) % 998244353;
    }
    long long ans = 0;
    for (int i = 0; i < n + 1; i++) {
      ans = (ans +
             ((i != 0 ? dp[i][0] : 0) + (i != n ? dp[i][1] : 0)) % 998244353) %
            998244353;
    }
    cout << ans << endl;
  }
  return 0;
}

```