#include <bits/stdc++.h>
using namespace std;
const long long int N = (long long int)(5e5 + 5);
const long long int inf = 2e18;
const long long int M1 = (long long int)(1e9 + 7);
const long long int M2 = (long long int)(998244353);
long long int n, m, a, b, c, d;
long long int w, h;
long long int x, y;
long long int t;
long long int dp[N][2];
void solve() {
  cin >> n;
  for (long long int i = 0; i <= n + 3; i++) {
    dp[i][0] = dp[i][1] = 0;
  }
  vector<long long int> arr;
  for (long long int i = 0; i < n; i++) {
    cin >> x;
    arr.push_back(x);
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    x = arr[i];
    if (x == 0) {
      ans += 1 + dp[0][0] + dp[2][1];
      dp[2][1] = (2 * dp[2][1]) % M2;
      dp[0][0] = (2 * dp[0][0] + 1) % M2;
    } else if (x == 1) {
      ans += dp[0][0] + dp[1][0] + dp[1][1] + dp[3][1] + 1;
      dp[3][1] = (2 * dp[3][1]) % M2;
      dp[1][1] = (2 * dp[1][1] + 1) % M2;
      dp[1][0] = (2 * dp[1][0] + dp[0][0]) % M2;
    } else {
      ans += dp[x - 2][0] + dp[x - 1][0] + dp[x][0] + dp[x][1] + dp[x + 2][1];
      dp[x + 2][1] = (2 * dp[x + 2][1]) % M2;
      dp[x][0] = (2 * dp[x][0] + dp[x - 1][0]) % M2;
      dp[x][1] = (dp[x - 2][0] + 2 * dp[x][1]) % M2;
    }
    ans = ans % M2;
  }
  cout << ans % M2 << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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;
const long long int N = (long long int)(5e5 + 5);
const long long int inf = 2e18;
const long long int M1 = (long long int)(1e9 + 7);
const long long int M2 = (long long int)(998244353);
long long int n, m, a, b, c, d;
long long int w, h;
long long int x, y;
long long int t;
long long int dp[N][2];
void solve() {
  cin >> n;
  for (long long int i = 0; i <= n + 3; i++) {
    dp[i][0] = dp[i][1] = 0;
  }
  vector<long long int> arr;
  for (long long int i = 0; i < n; i++) {
    cin >> x;
    arr.push_back(x);
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    x = arr[i];
    if (x == 0) {
      ans += 1 + dp[0][0] + dp[2][1];
      dp[2][1] = (2 * dp[2][1]) % M2;
      dp[0][0] = (2 * dp[0][0] + 1) % M2;
    } else if (x == 1) {
      ans += dp[0][0] + dp[1][0] + dp[1][1] + dp[3][1] + 1;
      dp[3][1] = (2 * dp[3][1]) % M2;
      dp[1][1] = (2 * dp[1][1] + 1) % M2;
      dp[1][0] = (2 * dp[1][0] + dp[0][0]) % M2;
    } else {
      ans += dp[x - 2][0] + dp[x - 1][0] + dp[x][0] + dp[x][1] + dp[x + 2][1];
      dp[x + 2][1] = (2 * dp[x + 2][1]) % M2;
      dp[x][0] = (2 * dp[x][0] + dp[x - 1][0]) % M2;
      dp[x][1] = (dp[x - 2][0] + 2 * dp[x][1]) % M2;
    }
    ans = ans % M2;
  }
  cout << ans % M2 << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```