#include <bits/stdc++.h>
using namespace std;
long long powerr[1000015];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  powerr[0] = 1;
  for (long long i = 1; i < 1000015; i++) {
    powerr[i] = powerr[i - 1] * 2;
    powerr[i] %= 998244353;
  }
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long freq_dp[n + 8];
    long long freq[n + 8];
    long long finax[n + 8];
    memset(finax, 0, sizeof(finax));
    memset(freq, 0, sizeof(freq));
    memset(freq_dp, 0, sizeof(freq_dp));
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long dp[n + 8];
    memset(dp, 0, sizeof(dp));
    for (long long i = n - 1; i >= 0; i--) {
      dp[i] = (freq_dp[a[i]] + freq_dp[a[i] + 1] + finax[a[i] + 2] + 1);
      dp[i] %= 998244353;
      freq_dp[a[i]] += dp[i];
      freq_dp[a[i]] %= 998244353;
      if (a[i] < 2) {
        finax[a[i]] += (powerr[freq[a[i]]]);
      }
      if (a[i] >= 2) {
        finax[a[i]] += (powerr[freq[a[i] - 2] + freq[a[i]]]);
      }
      finax[a[i]] += 998244353;
      finax[a[i]] %= 998244353;
      freq[a[i]]++;
    }
    long long ans = 0;
    ans = powerr[freq[1]] - 1;
    ans += 998244353;
    ans %= 998244353;
    for (long long i = 0; i < n; i++) {
      if (a[i] == 0) {
        ans += dp[i];
        ans %= 998244353;
      }
    }
    cout << ans << "\n";
  }
}

### Prompt

Please formulate 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 powerr[1000015];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  powerr[0] = 1;
  for (long long i = 1; i < 1000015; i++) {
    powerr[i] = powerr[i - 1] * 2;
    powerr[i] %= 998244353;
  }
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long freq_dp[n + 8];
    long long freq[n + 8];
    long long finax[n + 8];
    memset(finax, 0, sizeof(finax));
    memset(freq, 0, sizeof(freq));
    memset(freq_dp, 0, sizeof(freq_dp));
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long dp[n + 8];
    memset(dp, 0, sizeof(dp));
    for (long long i = n - 1; i >= 0; i--) {
      dp[i] = (freq_dp[a[i]] + freq_dp[a[i] + 1] + finax[a[i] + 2] + 1);
      dp[i] %= 998244353;
      freq_dp[a[i]] += dp[i];
      freq_dp[a[i]] %= 998244353;
      if (a[i] < 2) {
        finax[a[i]] += (powerr[freq[a[i]]]);
      }
      if (a[i] >= 2) {
        finax[a[i]] += (powerr[freq[a[i] - 2] + freq[a[i]]]);
      }
      finax[a[i]] += 998244353;
      finax[a[i]] %= 998244353;
      freq[a[i]]++;
    }
    long long ans = 0;
    ans = powerr[freq[1]] - 1;
    ans += 998244353;
    ans %= 998244353;
    for (long long i = 0; i < n; i++) {
      if (a[i] == 0) {
        ans += dp[i];
        ans %= 998244353;
      }
    }
    cout << ans << "\n";
  }
}

```