#include <bits/stdc++.h>
using namespace std;
long long int k = 998244353;
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % k;
  while (y > 0) {
    if (y & 1) res = (res * x) % k;
    y = y >> 1;
    x = (x * x) % k;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  cin >> test;
  while (test--) {
    int n;
    cin >> n;
    long long int a[n];
    vector<pair<long long int, long long int>> dp(n + 5, {0, 0});
    for (int i = 0; i < n; i++) cin >> a[i];
    dp[0].first = 1;
    for (int i = 0; i < n; i++) {
      long long int curr = a[i];
      curr--;
      if (curr >= 0) {
        long long int temp = dp[curr].first + dp[curr].second;
        temp %= k;
        dp[curr].second += temp;
        dp[curr].second %= k;
      }
      curr += 2;
      dp[curr].second += dp[curr].second;
      dp[curr].first += dp[curr].first;
      dp[curr].second %= k;
      dp[curr].first %= k;
      dp[curr].first += dp[curr - 1].first;
      dp[curr].first %= k;
    }
    long long int ans = 0;
    for (int i = 0; i <= n + 2; i++) {
      ans += dp[i].first;
      ans %= k;
      ans += dp[i].second;
      ans %= k;
    }
    ans--;
    ans %= k;
    if (ans < 0) ans += k;
    cout << ans;
    cout << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write 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 int k = 998244353;
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % k;
  while (y > 0) {
    if (y & 1) res = (res * x) % k;
    y = y >> 1;
    x = (x * x) % k;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  cin >> test;
  while (test--) {
    int n;
    cin >> n;
    long long int a[n];
    vector<pair<long long int, long long int>> dp(n + 5, {0, 0});
    for (int i = 0; i < n; i++) cin >> a[i];
    dp[0].first = 1;
    for (int i = 0; i < n; i++) {
      long long int curr = a[i];
      curr--;
      if (curr >= 0) {
        long long int temp = dp[curr].first + dp[curr].second;
        temp %= k;
        dp[curr].second += temp;
        dp[curr].second %= k;
      }
      curr += 2;
      dp[curr].second += dp[curr].second;
      dp[curr].first += dp[curr].first;
      dp[curr].second %= k;
      dp[curr].first %= k;
      dp[curr].first += dp[curr - 1].first;
      dp[curr].first %= k;
    }
    long long int ans = 0;
    for (int i = 0; i <= n + 2; i++) {
      ans += dp[i].first;
      ans %= k;
      ans += dp[i].second;
      ans %= k;
    }
    ans--;
    ans %= k;
    if (ans < 0) ans += k;
    cout << ans;
    cout << "\n";
  }
  return 0;
}

```