#include <bits/stdc++.h>
using namespace std;
long long md = 998244353;
long long modRecursive(long long a, long long k, long long md) {
  if (k == 0) return 1;
  long long temp = modRecursive((a * a) % md, k / 2, md) % md;
  if (k % 2) temp = (temp * a % md) % md;
  return temp;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    map<long long, long long> mp;
    long long maxx = 0;
    for (auto &it : a) {
      cin >> it;
      maxx = max(maxx, it);
    }
    long long curr = 0, ans = 0;
    vector<long long> dp(maxx + 2), extra(maxx + 2);
    for (long long i = 0; i < n; i++) {
      if (a[i] == 0) {
        dp[a[i]] = dp[a[i]] * 2 + 1;
        extra[a[i] + 1] *= 2;
      } else if (a[i] == 1) {
        dp[a[i]] = dp[a[i]] * 2 + dp[a[i] - 1];
        extra[a[i] + 1] *= 2;
        extra[a[i] - 1] = extra[a[i] - 1] * 2 + 1;
        extra[a[i] - 1] %= md;
      } else {
        dp[a[i]] = dp[a[i]] * 2 + dp[a[i] - 1];
        extra[a[i] + 1] *= 2;
        extra[a[i] - 1] = extra[a[i] - 1] * 2 + dp[a[i] - 2];
        extra[a[i] - 1] %= md;
      }
      dp[a[i]] %= md;
      extra[a[i] + 1] %= md;
      extra[a[i]] %= md;
    }
    for (long long i = 0; i <= maxx; i++) {
      ans = (ans + dp[i] + extra[i]) % md;
    }
    cout << (ans + md) % md << endl;
  }
}

### 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;
long long md = 998244353;
long long modRecursive(long long a, long long k, long long md) {
  if (k == 0) return 1;
  long long temp = modRecursive((a * a) % md, k / 2, md) % md;
  if (k % 2) temp = (temp * a % md) % md;
  return temp;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    map<long long, long long> mp;
    long long maxx = 0;
    for (auto &it : a) {
      cin >> it;
      maxx = max(maxx, it);
    }
    long long curr = 0, ans = 0;
    vector<long long> dp(maxx + 2), extra(maxx + 2);
    for (long long i = 0; i < n; i++) {
      if (a[i] == 0) {
        dp[a[i]] = dp[a[i]] * 2 + 1;
        extra[a[i] + 1] *= 2;
      } else if (a[i] == 1) {
        dp[a[i]] = dp[a[i]] * 2 + dp[a[i] - 1];
        extra[a[i] + 1] *= 2;
        extra[a[i] - 1] = extra[a[i] - 1] * 2 + 1;
        extra[a[i] - 1] %= md;
      } else {
        dp[a[i]] = dp[a[i]] * 2 + dp[a[i] - 1];
        extra[a[i] + 1] *= 2;
        extra[a[i] - 1] = extra[a[i] - 1] * 2 + dp[a[i] - 2];
        extra[a[i] - 1] %= md;
      }
      dp[a[i]] %= md;
      extra[a[i] + 1] %= md;
      extra[a[i]] %= md;
    }
    for (long long i = 0; i <= maxx; i++) {
      ans = (ans + dp[i] + extra[i]) % md;
    }
    cout << (ans + md) % md << endl;
  }
}

```