#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const long long inf = 1e18;
const long long MAXN = 2e5 + 1;
const long long R = 1 << 20;
const long long mod = 998244353;
const long long LOGN = 21;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (auto &x : a) cin >> x;
  vector<long long> COUNT(n + 5);
  vector<long long> pow2(n + 5, 1);
  for (long long i = 1; i < n + 5; ++i) (pow2[i] = pow2[i - 1] * 2) %= mod;
  for (long long i = 0; i < n; ++i) COUNT[a[i]]++;
  vector<long long> cnt(n + 5);
  long long ans = 0;
  for (long long i = 0; i < n; ++i) {
    COUNT[a[i]]--;
    if (a[i] == 0) {
      (cnt[0] += cnt[0] + 1) %= mod;
      continue;
    }
    if (a[i] == 1) {
      (cnt[1] += cnt[1] + cnt[0]) %= mod;
      (ans += pow2[COUNT[1]]) %= mod;
      continue;
    }
    long long x = a[i];
    (cnt[x] += cnt[x] + cnt[x - 1]) %= mod;
    (ans += pow2[COUNT[x] + COUNT[x - 2]] * cnt[x - 2]) %= mod;
  }
  for (long long i = 0; i <= n; ++i) (ans += cnt[i]) %= mod;
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long _ = 1;
  cin >> _;
  while (_--) solve();
  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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const long long inf = 1e18;
const long long MAXN = 2e5 + 1;
const long long R = 1 << 20;
const long long mod = 998244353;
const long long LOGN = 21;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (auto &x : a) cin >> x;
  vector<long long> COUNT(n + 5);
  vector<long long> pow2(n + 5, 1);
  for (long long i = 1; i < n + 5; ++i) (pow2[i] = pow2[i - 1] * 2) %= mod;
  for (long long i = 0; i < n; ++i) COUNT[a[i]]++;
  vector<long long> cnt(n + 5);
  long long ans = 0;
  for (long long i = 0; i < n; ++i) {
    COUNT[a[i]]--;
    if (a[i] == 0) {
      (cnt[0] += cnt[0] + 1) %= mod;
      continue;
    }
    if (a[i] == 1) {
      (cnt[1] += cnt[1] + cnt[0]) %= mod;
      (ans += pow2[COUNT[1]]) %= mod;
      continue;
    }
    long long x = a[i];
    (cnt[x] += cnt[x] + cnt[x - 1]) %= mod;
    (ans += pow2[COUNT[x] + COUNT[x - 2]] * cnt[x - 2]) %= mod;
  }
  for (long long i = 0; i <= n; ++i) (ans += cnt[i]) %= mod;
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long _ = 1;
  cin >> _;
  while (_--) solve();
  return 0;
}

```