#include <bits/stdc++.h>
using namespace std;
using LL = long long;
int cases = -1;
enum InputType : int {
  SingleInput = 1,
  MultipleInput = 2,
  NotSet = 0,
};
void getCases() {
  InputType inputType = MultipleInput;
  if (inputType == SingleInput) {
    cases = 1;
  } else if (inputType == MultipleInput) {
    scanf("%d", &cases);
  } else {
    assert(false && "fix the function");
  }
}
int n;
int ar[500005];
LL dp[2][500005];
const LL MOD = 998244353;
void Input() {
  scanf("%d", &n);
  for (int i = 0; i <= n + 3; ++i) {
    dp[0][i] = dp[1][i] = 0;
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", &ar[i]);
  }
}
LL add(LL x, LL y) { return (x + y) % MOD; }
int main() {
  getCases();
  while (cases--) {
    Input();
    dp[0][0] = 1;
    for (int i = 0; i < n; ++i) {
      int x = ar[i];
      dp[0][x + 1] = add(dp[0][x + 1], dp[0][x + 1]);
      dp[0][x + 1] = add(dp[0][x + 1], dp[0][x]);
      if (x > 0) {
        dp[1][x - 1] = add(dp[1][x - 1], dp[1][x - 1]);
        dp[1][x - 1] = add(dp[1][x - 1], dp[0][x - 1]);
      }
      dp[1][x + 1] = add(dp[1][x + 1], dp[1][x + 1]);
    }
    LL ans = 0;
    for (int i = 0; i <= n + 3; ++i) {
      ans = add(ans, dp[0][i]);
      ans = add(ans, dp[1][i]);
    }
    ans = add(ans, MOD - 1);
    printf("%lld\n", ans);
  }
}

### Prompt

In CPP, your task is to solve 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;
int cases = -1;
enum InputType : int {
  SingleInput = 1,
  MultipleInput = 2,
  NotSet = 0,
};
void getCases() {
  InputType inputType = MultipleInput;
  if (inputType == SingleInput) {
    cases = 1;
  } else if (inputType == MultipleInput) {
    scanf("%d", &cases);
  } else {
    assert(false && "fix the function");
  }
}
int n;
int ar[500005];
LL dp[2][500005];
const LL MOD = 998244353;
void Input() {
  scanf("%d", &n);
  for (int i = 0; i <= n + 3; ++i) {
    dp[0][i] = dp[1][i] = 0;
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", &ar[i]);
  }
}
LL add(LL x, LL y) { return (x + y) % MOD; }
int main() {
  getCases();
  while (cases--) {
    Input();
    dp[0][0] = 1;
    for (int i = 0; i < n; ++i) {
      int x = ar[i];
      dp[0][x + 1] = add(dp[0][x + 1], dp[0][x + 1]);
      dp[0][x + 1] = add(dp[0][x + 1], dp[0][x]);
      if (x > 0) {
        dp[1][x - 1] = add(dp[1][x - 1], dp[1][x - 1]);
        dp[1][x - 1] = add(dp[1][x - 1], dp[0][x - 1]);
      }
      dp[1][x + 1] = add(dp[1][x + 1], dp[1][x + 1]);
    }
    LL ans = 0;
    for (int i = 0; i <= n + 3; ++i) {
      ans = add(ans, dp[0][i]);
      ans = add(ans, dp[1][i]);
    }
    ans = add(ans, MOD - 1);
    printf("%lld\n", ans);
  }
}

```