#include <bits/stdc++.h>
using namespace std;
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
constexpr int MAXN = 5e5 + 2;
constexpr int MOD = 998244353;
constexpr long long INF = 1e18 + 10;
constexpr int LOG = 3e1 + 0;
constexpr int SQ = 2e3 + 20;
constexpr int ALP = 130;
int tc, n, a[MAXN], dp[MAXN][2];
int add(int a, int b) { return (a + b) % MOD; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  while (tc--) {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (int i = 1; i <= n; i++) {
      dp[a[i] + 2][1] = add(dp[a[i] + 2][1], dp[a[i] + 2][1]);
      if (a[i] == 0) {
        dp[a[i]][0] = add(dp[a[i]][0] * 2 % MOD, 1);
      } else if (a[i] == 1) {
        dp[a[i]][1] = add(dp[a[i]][1] * 2 % MOD, 1);
        dp[a[i]][0] = add(dp[a[i]][0] * 2 % MOD, dp[a[i] - 1][0]);
      } else {
        dp[a[i]][0] = add(dp[a[i]][0] * 2 % MOD, dp[a[i] - 1][0]);
        dp[a[i]][1] = add(dp[a[i]][1] * 2 % MOD, dp[a[i] - 2][0]);
      }
    }
    int ans = 0;
    for (int i = 0; i <= n; i++) {
      ans = add(ans, dp[i][0]);
      ans = add(ans, dp[i][1]);
      dp[i][0] = 0;
      dp[i][1] = 0;
    }
    cout << ans << "\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;
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
constexpr int MAXN = 5e5 + 2;
constexpr int MOD = 998244353;
constexpr long long INF = 1e18 + 10;
constexpr int LOG = 3e1 + 0;
constexpr int SQ = 2e3 + 20;
constexpr int ALP = 130;
int tc, n, a[MAXN], dp[MAXN][2];
int add(int a, int b) { return (a + b) % MOD; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  while (tc--) {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (int i = 1; i <= n; i++) {
      dp[a[i] + 2][1] = add(dp[a[i] + 2][1], dp[a[i] + 2][1]);
      if (a[i] == 0) {
        dp[a[i]][0] = add(dp[a[i]][0] * 2 % MOD, 1);
      } else if (a[i] == 1) {
        dp[a[i]][1] = add(dp[a[i]][1] * 2 % MOD, 1);
        dp[a[i]][0] = add(dp[a[i]][0] * 2 % MOD, dp[a[i] - 1][0]);
      } else {
        dp[a[i]][0] = add(dp[a[i]][0] * 2 % MOD, dp[a[i] - 1][0]);
        dp[a[i]][1] = add(dp[a[i]][1] * 2 % MOD, dp[a[i] - 2][0]);
      }
    }
    int ans = 0;
    for (int i = 0; i <= n; i++) {
      ans = add(ans, dp[i][0]);
      ans = add(ans, dp[i][1]);
      dp[i][0] = 0;
      dp[i][1] = 0;
    }
    cout << ans << "\n";
  }
  return 0;
}

```