#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    vector<vector<long long>> dp(n + 6, vector<long long>(2, 0));
    set<long long> s;
    for (long long i = 0; i < n; i++) {
      long long x;
      cin >> x;
      a[i] = x;
      s.insert(i);
    }
    s.insert(n);
    s.insert(n + 1);
    dp[0][0] = 0;
    long long ans = 0;
    long long last = 0;
    for (long long i = 0; i < n; i++) {
      last = a[i];
      if ((a[i] == 0)) {
        dp[1][0] = (2 * dp[1][0]) % MOD + 1LL;
        dp[1][0] = dp[1][0] % MOD;
        dp[1][1] = (2 * dp[1][1]) % MOD;
      } else if (a[i] == 1) {
        dp[0][1] = (2 * dp[0][1]) % MOD + 1LL;
        dp[0][1] = dp[0][1] % MOD;
        dp[2][0] = ((2 * dp[2][0]) % MOD + (dp[1][0] % MOD)) % MOD;
        dp[2][0] = (dp[2][0]) % MOD;
        dp[2][1] = (2 * dp[2][1]) % MOD;
      } else {
        dp[a[i] - 1][1] =
            ((2 * dp[a[i] - 1][1]) % MOD + (dp[a[i] - 1][0]) % MOD) % MOD;
        dp[a[i] + 1][0] =
            ((2 * dp[a[i] + 1][0]) % MOD + (dp[a[i]][0] % MOD)) % MOD;
        dp[a[i] + 1][1] = (2 * dp[a[i] + 1][1]) % MOD;
      }
    }
    for (auto &i : s) {
      auto val = i;
      last = val;
      ans = (ans + dp[val][0] % MOD) % MOD;
      ans = (ans + dp[val][1] % MOD) % MOD;
      ans = (ans) % MOD;
    }
    cout << ans << "\n";
  }
}

### 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 long long MOD = 998244353;
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    vector<vector<long long>> dp(n + 6, vector<long long>(2, 0));
    set<long long> s;
    for (long long i = 0; i < n; i++) {
      long long x;
      cin >> x;
      a[i] = x;
      s.insert(i);
    }
    s.insert(n);
    s.insert(n + 1);
    dp[0][0] = 0;
    long long ans = 0;
    long long last = 0;
    for (long long i = 0; i < n; i++) {
      last = a[i];
      if ((a[i] == 0)) {
        dp[1][0] = (2 * dp[1][0]) % MOD + 1LL;
        dp[1][0] = dp[1][0] % MOD;
        dp[1][1] = (2 * dp[1][1]) % MOD;
      } else if (a[i] == 1) {
        dp[0][1] = (2 * dp[0][1]) % MOD + 1LL;
        dp[0][1] = dp[0][1] % MOD;
        dp[2][0] = ((2 * dp[2][0]) % MOD + (dp[1][0] % MOD)) % MOD;
        dp[2][0] = (dp[2][0]) % MOD;
        dp[2][1] = (2 * dp[2][1]) % MOD;
      } else {
        dp[a[i] - 1][1] =
            ((2 * dp[a[i] - 1][1]) % MOD + (dp[a[i] - 1][0]) % MOD) % MOD;
        dp[a[i] + 1][0] =
            ((2 * dp[a[i] + 1][0]) % MOD + (dp[a[i]][0] % MOD)) % MOD;
        dp[a[i] + 1][1] = (2 * dp[a[i] + 1][1]) % MOD;
      }
    }
    for (auto &i : s) {
      auto val = i;
      last = val;
      ans = (ans + dp[val][0] % MOD) % MOD;
      ans = (ans + dp[val][1] % MOD) % MOD;
      ans = (ans) % MOD;
    }
    cout << ans << "\n";
  }
}

```