#include <bits/stdc++.h>
using namespace std;
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
template <typename T, typename... Args>
inline string arrStr(T arr, int n) {
  stringstream s;
  s << "[";
  for (int i = 0; i < n - 1; i++) s << arr[i] << ",";
  s << arr[n - 1] << "]";
  return s.str();
}
inline void __evars_begin(int line) { cerr << "#" << line << ": "; }
template <typename T>
inline void __evars_out_var(vector<T> val) {
  cerr << arrStr(val, val.size());
}
template <typename T>
inline void __evars_out_var(T* val) {
  cerr << arrStr(val, 10);
}
template <typename T>
inline void __evars_out_var(T val) {
  cerr << val;
}
inline void __evars(vector<string>::iterator it) { cerr << endl; }
template <typename T, typename... Args>
inline void __evars(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << "=";
  __evars_out_var(a);
  cerr << "; ";
  __evars(++it, args...);
}
const long long N = 2e5 + 15;
const long long sz = (1 << 20);
const long long mod = 1e9 + 7;
const long long mod2 = 998244353;
void boomer() {
  long long n;
  cin >> n;
  vector<long long> v(n + 1);
  vector<long long> a;
  for (long long i = 1; i <= n; ++i) {
    cin >> v[i];
    a.push_back(v[i]);
  }
  sort(a.begin(), a.end());
  long long mex = 0;
  for (long long x : a) {
    if (x == mex) mex++;
  }
  vector<long long> dp1(mex + 10), dp2(mex + 10);
  for (long long i = 1; i <= n; ++i) {
    if (v[i]) {
      long long x = v[i] - 1;
      dp2[x] = (dp2[x] + dp2[x] + dp1[x]) % mod2;
    }
    long long x = v[i] + 1;
    dp1[x] = (dp1[x] + dp1[x]) % mod2;
    dp2[x] = (dp2[x] + dp2[x]) % mod2;
    dp1[x] = (dp1[x] + dp1[x - 1]) % mod2;
    if (v[i] == 0) dp1[1]++;
    if (v[i] == 1) dp2[0]++;
  }
  long long ans = 0;
  for (long long i = 0; i <= mex; ++i) ans = (ans + dp1[i] + dp2[i]) % mod2;
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long tc = 1;
  cin >> tc;
  for (long long i = 1; i <= tc; ++i) {
    boomer();
    if (i < tc) cout << "\n";
  }
  return 0;
}

### 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;
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
template <typename T, typename... Args>
inline string arrStr(T arr, int n) {
  stringstream s;
  s << "[";
  for (int i = 0; i < n - 1; i++) s << arr[i] << ",";
  s << arr[n - 1] << "]";
  return s.str();
}
inline void __evars_begin(int line) { cerr << "#" << line << ": "; }
template <typename T>
inline void __evars_out_var(vector<T> val) {
  cerr << arrStr(val, val.size());
}
template <typename T>
inline void __evars_out_var(T* val) {
  cerr << arrStr(val, 10);
}
template <typename T>
inline void __evars_out_var(T val) {
  cerr << val;
}
inline void __evars(vector<string>::iterator it) { cerr << endl; }
template <typename T, typename... Args>
inline void __evars(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << "=";
  __evars_out_var(a);
  cerr << "; ";
  __evars(++it, args...);
}
const long long N = 2e5 + 15;
const long long sz = (1 << 20);
const long long mod = 1e9 + 7;
const long long mod2 = 998244353;
void boomer() {
  long long n;
  cin >> n;
  vector<long long> v(n + 1);
  vector<long long> a;
  for (long long i = 1; i <= n; ++i) {
    cin >> v[i];
    a.push_back(v[i]);
  }
  sort(a.begin(), a.end());
  long long mex = 0;
  for (long long x : a) {
    if (x == mex) mex++;
  }
  vector<long long> dp1(mex + 10), dp2(mex + 10);
  for (long long i = 1; i <= n; ++i) {
    if (v[i]) {
      long long x = v[i] - 1;
      dp2[x] = (dp2[x] + dp2[x] + dp1[x]) % mod2;
    }
    long long x = v[i] + 1;
    dp1[x] = (dp1[x] + dp1[x]) % mod2;
    dp2[x] = (dp2[x] + dp2[x]) % mod2;
    dp1[x] = (dp1[x] + dp1[x - 1]) % mod2;
    if (v[i] == 0) dp1[1]++;
    if (v[i] == 1) dp2[0]++;
  }
  long long ans = 0;
  for (long long i = 0; i <= mex; ++i) ans = (ans + dp1[i] + dp2[i]) % mod2;
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long tc = 1;
  cin >> tc;
  for (long long i = 1; i <= tc; ++i) {
    boomer();
    if (i < tc) cout << "\n";
  }
  return 0;
}

```