#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int mod = 998244353;
const double PI = 3.14159265359;
const double eps = 1e-9;
int n, v[500002], dp[2][500005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (; t; --t) {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> v[i];
    int ans = 0;
    dp[1][0] = 1;
    for (int i = 1; i <= n; ++i) {
      dp[1][v[i] + 1] += dp[1][v[i] + 1];
      if (dp[1][v[i] + 1] >= mod) dp[1][v[i] + 1] -= mod;
      dp[1][v[i] + 1] += dp[1][v[i]];
      if (dp[1][v[i] + 1] >= mod) dp[1][v[i] + 1] -= mod;
      if (v[i] > 0) {
        dp[0][v[i] - 1] += dp[0][v[i] - 1];
        if (dp[0][v[i] - 1] >= mod) dp[0][v[i] - 1] -= mod;
        dp[0][v[i] - 1] += dp[1][v[i] - 1];
        if (dp[0][v[i] - 1] >= mod) dp[0][v[i] - 1] -= mod;
      }
      dp[0][v[i] + 1] += dp[0][v[i] + 1];
      if (dp[0][v[i] + 1] >= mod) dp[0][v[i] + 1] -= mod;
    }
    for (int i = 0; i <= n + 2; ++i) {
      ans = ans + dp[0][i];
      while (ans >= mod) ans -= mod;
      ans = ans + dp[1][i];
      while (ans >= mod) ans -= mod;
      dp[0][i] = dp[1][i] = 0;
    }
    cout << (ans + mod - 1) % mod << '\n';
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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>
#pragma GCC optimize("O3")
using namespace std;
const int mod = 998244353;
const double PI = 3.14159265359;
const double eps = 1e-9;
int n, v[500002], dp[2][500005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (; t; --t) {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> v[i];
    int ans = 0;
    dp[1][0] = 1;
    for (int i = 1; i <= n; ++i) {
      dp[1][v[i] + 1] += dp[1][v[i] + 1];
      if (dp[1][v[i] + 1] >= mod) dp[1][v[i] + 1] -= mod;
      dp[1][v[i] + 1] += dp[1][v[i]];
      if (dp[1][v[i] + 1] >= mod) dp[1][v[i] + 1] -= mod;
      if (v[i] > 0) {
        dp[0][v[i] - 1] += dp[0][v[i] - 1];
        if (dp[0][v[i] - 1] >= mod) dp[0][v[i] - 1] -= mod;
        dp[0][v[i] - 1] += dp[1][v[i] - 1];
        if (dp[0][v[i] - 1] >= mod) dp[0][v[i] - 1] -= mod;
      }
      dp[0][v[i] + 1] += dp[0][v[i] + 1];
      if (dp[0][v[i] + 1] >= mod) dp[0][v[i] + 1] -= mod;
    }
    for (int i = 0; i <= n + 2; ++i) {
      ans = ans + dp[0][i];
      while (ans >= mod) ans -= mod;
      ans = ans + dp[1][i];
      while (ans >= mod) ans -= mod;
      dp[0][i] = dp[1][i] = 0;
    }
    cout << (ans + mod - 1) % mod << '\n';
  }
  return 0;
}

```