#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using vi = vector<int>;
using vl = vector<ll>;
template <typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
const ll MOD = 998244353;
int n;
ll a[500005];
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  vl dp(n + 5, 0), dp2(n + 5, 0);
  dp[0] = 1;
  for (int i = 0; i < n; i++) {
    (dp[a[i] + 1] *= 2) %= MOD;
    (dp[a[i] + 1] += dp[a[i]]) %= MOD;
    if (a[i] > 0) {
      (dp2[a[i] - 1] *= 2) %= MOD;
      (dp2[a[i] - 1] += dp[a[i] - 1]) %= MOD;
    }
    (dp2[a[i] + 1] *= 2) %= MOD;
  }
  ll ans = 0;
  for (auto& l : dp) (ans += l) %= MOD;
  for (auto& l : dp2) (ans += l) %= MOD;
  cout << (ans - 1 + MOD) % MOD << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

### 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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using vi = vector<int>;
using vl = vector<ll>;
template <typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
const ll MOD = 998244353;
int n;
ll a[500005];
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  vl dp(n + 5, 0), dp2(n + 5, 0);
  dp[0] = 1;
  for (int i = 0; i < n; i++) {
    (dp[a[i] + 1] *= 2) %= MOD;
    (dp[a[i] + 1] += dp[a[i]]) %= MOD;
    if (a[i] > 0) {
      (dp2[a[i] - 1] *= 2) %= MOD;
      (dp2[a[i] - 1] += dp[a[i] - 1]) %= MOD;
    }
    (dp2[a[i] + 1] *= 2) %= MOD;
  }
  ll ans = 0;
  for (auto& l : dp) (ans += l) %= MOD;
  for (auto& l : dp2) (ans += l) %= MOD;
  cout << (ans - 1 + MOD) % MOD << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

```