#include <bits/stdc++.h>
using namespace std;
const long long int mma = 1e6 + 10;
const long long int mod = 998244353;
vector<vector<long long int>> p;
long long int n, m;
bool solve(long long int x) {
  vector<bool> c(m);
  bool check = 0;
  for (int i = 0; i < n; i++) {
    long long int cf = 0;
    for (int j = 0; j < m; j++) {
      if (p[i][j] >= x) {
        cf++;
        c[j] = 1;
      }
    }
    if (cf > 1) check = 1;
  }
  if (!check) {
    return 0;
  }
  bool checkagain = 0;
  for (bool b : c)
    if (!b) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int dp1[n + 2], dp2[n + 2];
    fill(dp1, dp1 + n + 2, 0);
    fill(dp2, dp2 + n + 2, 0);
    dp1[0] = 1;
    long long int ans = -1;
    for (int j = 0; j < n; j++) {
      long long int x;
      cin >> x;
      dp1[x + 1] = (2 * (dp1[x + 1])) % mod;
      dp1[x + 1] = (dp1[x + 1] + dp1[x]) % mod;
      if (x > 0) {
        dp2[x - 1] = (2 * (dp2[x - 1])) % mod;
        dp2[x - 1] = (dp2[x - 1] + dp1[x - 1]) % mod;
      }
      dp2[x + 1] = (2 * (dp2[x + 1])) % mod;
    }
    for (int i = 0; i < n + 2; i++) {
      ans += dp1[i];
      ans %= mod;
      ans += dp2[i];
      ans %= mod;
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 int mma = 1e6 + 10;
const long long int mod = 998244353;
vector<vector<long long int>> p;
long long int n, m;
bool solve(long long int x) {
  vector<bool> c(m);
  bool check = 0;
  for (int i = 0; i < n; i++) {
    long long int cf = 0;
    for (int j = 0; j < m; j++) {
      if (p[i][j] >= x) {
        cf++;
        c[j] = 1;
      }
    }
    if (cf > 1) check = 1;
  }
  if (!check) {
    return 0;
  }
  bool checkagain = 0;
  for (bool b : c)
    if (!b) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int dp1[n + 2], dp2[n + 2];
    fill(dp1, dp1 + n + 2, 0);
    fill(dp2, dp2 + n + 2, 0);
    dp1[0] = 1;
    long long int ans = -1;
    for (int j = 0; j < n; j++) {
      long long int x;
      cin >> x;
      dp1[x + 1] = (2 * (dp1[x + 1])) % mod;
      dp1[x + 1] = (dp1[x + 1] + dp1[x]) % mod;
      if (x > 0) {
        dp2[x - 1] = (2 * (dp2[x - 1])) % mod;
        dp2[x - 1] = (dp2[x - 1] + dp1[x - 1]) % mod;
      }
      dp2[x + 1] = (2 * (dp2[x + 1])) % mod;
    }
    for (int i = 0; i < n + 2; i++) {
      ans += dp1[i];
      ans %= mod;
      ans += dp2[i];
      ans %= mod;
    }
    cout << ans << "\n";
  }
  return 0;
}

```