#include <bits/stdc++.h>
using namespace std;
long long n, m, T, cnt;
long long ans;
string s;
vector<long long> v;
long long f1[1000010], f2[1000010];
const long long M = 998244353;
inline long long read() {
  char ch = getchar();
  long long ret = 0;
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();
  return ret;
}
signed main() {
  T = read();
  while (T--) {
    n = read();
    for (long long i = 0; i <= n; i++) f1[i] = f2[i] = 0;
    f1[0] = 1;
    for (long long i = 1; i <= n; i++) {
      long long x = read();
      f1[x + 1] = (f1[x + 1] + f1[x] + f1[x + 1]) % M;
      if (x > 0) {
        f2[x - 1] = (f2[x - 1] + f2[x - 1] + f1[x - 1]) % M;
      }
      f2[x + 1] = 2 * f2[x + 1] % M;
    }
    ans = 0;
    for (long long i = 0; i <= n; i++) {
      ans = (ans + f1[i] + f2[i]) % M;
    }
    printf("%lld\n", (ans - 1 + M) % M);
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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, m, T, cnt;
long long ans;
string s;
vector<long long> v;
long long f1[1000010], f2[1000010];
const long long M = 998244353;
inline long long read() {
  char ch = getchar();
  long long ret = 0;
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();
  return ret;
}
signed main() {
  T = read();
  while (T--) {
    n = read();
    for (long long i = 0; i <= n; i++) f1[i] = f2[i] = 0;
    f1[0] = 1;
    for (long long i = 1; i <= n; i++) {
      long long x = read();
      f1[x + 1] = (f1[x + 1] + f1[x] + f1[x + 1]) % M;
      if (x > 0) {
        f2[x - 1] = (f2[x - 1] + f2[x - 1] + f1[x - 1]) % M;
      }
      f2[x + 1] = 2 * f2[x + 1] % M;
    }
    ans = 0;
    for (long long i = 0; i <= n; i++) {
      ans = (ans + f1[i] + f2[i]) % M;
    }
    printf("%lld\n", (ans - 1 + M) % M);
  }
  return 0;
}

```