#include <bits/stdc++.h>
using namespace std;
const int N = 501234, M = 998244353;
int dp[N][3], sum[N][3], a[N];
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n + 10; i++) {
    for (int j = 0; j < 3; j++) {
      dp[i][j] = 0;
      sum[i][j] = 0;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j < 3; j++) {
      if (j == 0) {
        dp[i][j] = 1;
        dp[i][j] += sum[a[i] + 2][2];
        dp[i][j] += sum[a[i]][0];
        dp[i][j] %= M;
        sum[a[i]][j] += dp[i][j];
        sum[a[i]][j] %= M;
      } else if (j == 1) {
        dp[i][j] = 1;
        dp[i][j] += sum[a[i]][1];
        dp[i][j] += sum[a[i] + 1][1];
        dp[i][j] %= M;
        dp[i][j] += sum[a[i] + 2][2];
        dp[i][j] %= M;
        sum[a[i]][j] += dp[i][j];
        sum[a[i]][j] %= M;
      } else {
        dp[i][j] = 1;
        if (a[i] > 1) {
          dp[i][j] += sum[a[i] - 2][0];
        }
        dp[i][j] += sum[a[i]][2];
        dp[i][j] %= M;
        sum[a[i]][j] += dp[i][j];
        sum[a[i]][j] %= M;
      }
    }
  }
  int ans = sum[0][1] + sum[1][2];
  ans %= M;
  ans += M;
  ans %= M;
  cout << ans << ' ';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Please create a solution in cpp 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 int N = 501234, M = 998244353;
int dp[N][3], sum[N][3], a[N];
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n + 10; i++) {
    for (int j = 0; j < 3; j++) {
      dp[i][j] = 0;
      sum[i][j] = 0;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j < 3; j++) {
      if (j == 0) {
        dp[i][j] = 1;
        dp[i][j] += sum[a[i] + 2][2];
        dp[i][j] += sum[a[i]][0];
        dp[i][j] %= M;
        sum[a[i]][j] += dp[i][j];
        sum[a[i]][j] %= M;
      } else if (j == 1) {
        dp[i][j] = 1;
        dp[i][j] += sum[a[i]][1];
        dp[i][j] += sum[a[i] + 1][1];
        dp[i][j] %= M;
        dp[i][j] += sum[a[i] + 2][2];
        dp[i][j] %= M;
        sum[a[i]][j] += dp[i][j];
        sum[a[i]][j] %= M;
      } else {
        dp[i][j] = 1;
        if (a[i] > 1) {
          dp[i][j] += sum[a[i] - 2][0];
        }
        dp[i][j] += sum[a[i]][2];
        dp[i][j] %= M;
        sum[a[i]][j] += dp[i][j];
        sum[a[i]][j] %= M;
      }
    }
  }
  int ans = sum[0][1] + sum[1][2];
  ans %= M;
  ans += M;
  ans %= M;
  cout << ans << ' ';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```