#include <bits/stdc++.h>
using namespace std;
const int inf = 0x7f7f7f7f;
const long long linf = 1e18;
const int maxn = 5e5 + 9;
const int maxm = 1e5 + 9;
const double PI = 3.1415926;
const double eps = 1e-5;
const int base = 131;
const int N = 1e6;
const long long mod = 998244353;
namespace IO {
long long read() {
  long long a = 1, b = 0;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') a = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    b = (b << 3) + (b << 1) + c - '0';
    c = getchar();
  }
  return a * b;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
}  // namespace IO
using namespace IO;
int n;
long long dp[maxn][2];
int main() {
  int cse = read();
  while (cse--) {
    dp[0][0] = 1;
    n = read();
    for (int i = 1; i <= n; ++i) {
      int x = read() + 1;
      dp[x][0] += dp[x - 1][0] + dp[x][0];
      dp[x][1] += dp[x][1];
      if (x > 1) dp[x][1] += dp[x - 2][0];
      dp[x + 2][1] += dp[x + 2][1];
      dp[x][0] %= mod;
      dp[x][1] %= mod;
      dp[x + 2][1] %= mod;
    }
    long long ans = 0;
    for (int i = 1; i <= n + 1; ++i) {
      ans = (ans + dp[i][0] + dp[i][1]) % mod;
    }
    print(ans);
    putchar('\n');
    for (int i = 0; i <= n + 1; ++i) dp[i][0] = dp[i][1] = 0;
  }
  return 0;
}

### Prompt

Your task is to create 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;
const int inf = 0x7f7f7f7f;
const long long linf = 1e18;
const int maxn = 5e5 + 9;
const int maxm = 1e5 + 9;
const double PI = 3.1415926;
const double eps = 1e-5;
const int base = 131;
const int N = 1e6;
const long long mod = 998244353;
namespace IO {
long long read() {
  long long a = 1, b = 0;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') a = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    b = (b << 3) + (b << 1) + c - '0';
    c = getchar();
  }
  return a * b;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
}  // namespace IO
using namespace IO;
int n;
long long dp[maxn][2];
int main() {
  int cse = read();
  while (cse--) {
    dp[0][0] = 1;
    n = read();
    for (int i = 1; i <= n; ++i) {
      int x = read() + 1;
      dp[x][0] += dp[x - 1][0] + dp[x][0];
      dp[x][1] += dp[x][1];
      if (x > 1) dp[x][1] += dp[x - 2][0];
      dp[x + 2][1] += dp[x + 2][1];
      dp[x][0] %= mod;
      dp[x][1] %= mod;
      dp[x + 2][1] %= mod;
    }
    long long ans = 0;
    for (int i = 1; i <= n + 1; ++i) {
      ans = (ans + dp[i][0] + dp[i][1]) % mod;
    }
    print(ans);
    putchar('\n');
    for (int i = 0; i <= n + 1; ++i) dp[i][0] = dp[i][1] = 0;
  }
  return 0;
}

```