#include <bits/stdc++.h>
using namespace std;
long long N, A[1000001], Dp[1000001], E[1000001];
inline long long Read() {
  register long long Return(0), Flag(0);
  register char C(getchar());
  for (; !isdigit(C); C = getchar()) Flag ^= !(C ^ 45);
  for (; isdigit(C); C = getchar())
    Return = (Return << 1) + (Return << 3) + (C ^ '0');
  if (Flag) Return = -Return;
  return Return;
}
inline void Write(long long X) {
  if (X < 0) X = -X, putchar('-');
  if (X > 9) Write(X / 10);
  putchar(X % 10 ^ 48);
}
signed main(void) {
  ios::sync_with_stdio(false);
  register long long Case(Read());
  while (Case--) {
    register long long i;
    N = Read();
    for (i = 1; i <= N + 5; i++) Dp[i] = E[i] = 0;
    Dp[0] = 1;
    for (i = 1; i <= N; i++) A[i] = Read();
    register long long Ret(0);
    for (i = 1; i <= N; i++) {
      Ret += E[A[i]] + E[A[i] + 2];
      if (A[i]) Ret += Dp[A[i] - 1];
      Ret %= 998244353;
      E[A[i]] = 2 * E[A[i]];
      E[A[i] + 2] = 2 * E[A[i] + 2];
      if (A[i]) E[A[i]] += Dp[A[i] - 1];
      E[A[i]] %= 998244353;
      E[A[i] + 2] %= 998244353;
      if (!A[i])
        Ret += Dp[0] + Dp[1], Ret %= 998244353;
      else
        Ret += Dp[A[i] + 1] + Dp[A[i]], Ret %= 998244353;
      Dp[A[i] + 1] = Dp[A[i] + 1] * 2 + Dp[A[i]], Dp[A[i] + 1] %= 998244353;
    }
    Write(Ret), putchar('\n');
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 N, A[1000001], Dp[1000001], E[1000001];
inline long long Read() {
  register long long Return(0), Flag(0);
  register char C(getchar());
  for (; !isdigit(C); C = getchar()) Flag ^= !(C ^ 45);
  for (; isdigit(C); C = getchar())
    Return = (Return << 1) + (Return << 3) + (C ^ '0');
  if (Flag) Return = -Return;
  return Return;
}
inline void Write(long long X) {
  if (X < 0) X = -X, putchar('-');
  if (X > 9) Write(X / 10);
  putchar(X % 10 ^ 48);
}
signed main(void) {
  ios::sync_with_stdio(false);
  register long long Case(Read());
  while (Case--) {
    register long long i;
    N = Read();
    for (i = 1; i <= N + 5; i++) Dp[i] = E[i] = 0;
    Dp[0] = 1;
    for (i = 1; i <= N; i++) A[i] = Read();
    register long long Ret(0);
    for (i = 1; i <= N; i++) {
      Ret += E[A[i]] + E[A[i] + 2];
      if (A[i]) Ret += Dp[A[i] - 1];
      Ret %= 998244353;
      E[A[i]] = 2 * E[A[i]];
      E[A[i] + 2] = 2 * E[A[i] + 2];
      if (A[i]) E[A[i]] += Dp[A[i] - 1];
      E[A[i]] %= 998244353;
      E[A[i] + 2] %= 998244353;
      if (!A[i])
        Ret += Dp[0] + Dp[1], Ret %= 998244353;
      else
        Ret += Dp[A[i] + 1] + Dp[A[i]], Ret %= 998244353;
      Dp[A[i] + 1] = Dp[A[i] + 1] * 2 + Dp[A[i]], Dp[A[i] + 1] %= 998244353;
    }
    Write(Ret), putchar('\n');
  }
  return 0;
}

```