#include <bits/stdc++.h>
using namespace std;
using namespace chrono;
template <class T>
void show(vector<vector<T>> &a) {
  for (auto &i : a) {
    for (T &j : i) cout << j << ' ';
    cout << endl;
  }
}
template <class T>
void show(vector<T> &a) {
  for (T i : a) cout << i << ' ';
  cout << endl;
}
template <class T>
void show(vector<pair<T, T>> &a) {
  for (auto &i : a) {
    cout << i.first << ' ' << i.second << endl;
  }
}
template <class T>
void show(pair<T, T> p) {
  cout << p.first << ' ' << p.second << endl;
}
void show(vector<string> &a) {
  for (auto &i : a) cout << i << endl;
}
template <class T>
void take(vector<T> &a) {
  for (auto &i : a) cin >> i;
}
long long digits(long long n) {
  long long k = 0;
  while (n) k++, n /= 10;
  return k;
}
bool isprime(long long n) {
  long long i;
  if (n == 1) return 0;
  if (n == 2) return 1;
  if (n % 2 == 0) return 0;
  for (i = 3; i * i <= n; i += 2) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long power(long long x, long long y, long long MOD) {
  long long ans = 1;
  x %= MOD;
  while (y > 0) {
    if (y % 2 == 1) ans = (ans * x) % MOD;
    y /= 2;
    x = (x * x) % MOD;
  }
  return ans;
}
void sieve(vector<bool> &a, long long n) {
  long long i, j;
  a[1] = 0;
  a[0] = 0;
  long long h = sqrt(n + 1);
  for (i = 2; i < h; i++) {
    if (!a[i]) continue;
    for (j = 2 * i; j < n + 1; j += i) a[j] = 0;
  }
}
void load(unordered_map<long long, long long> &g) { g.max_load_factor(0.25); }
vector<long long> bits(long long n) {
  vector<long long> a;
  long long i;
  for (i = 0; i < 33; i++) {
    a.push_back(n % 2);
    n /= 2;
  }
  reverse(a.begin(), a.end());
  return a;
}
void sub() {}
const long long N = 1e9 + 7;
const long long M = 998244353;
void solve() {
  long long i, j, n, m;
  cin >> n;
  vector<long long> a(n);
  take(a);
  long long ans = 0;
  for (i = 0; i < n; i++)
    if (a[i] == 1) ans++;
  ans = (power(2, ans, M) - 1 + M) % M;
  vector<long long> b(n + 3), c(n + 3);
  for (i = 0; i < n; i++) {
    if (a[i] == 0) b[0] = (2 * b[0] + 1) % M;
    if (a[i] == 1) b[1] = (2 * b[1] + b[0]) % M;
    if (a[i] > 1) b[a[i]] = (2 * b[a[i]] + b[a[i] - 1]) % M;
    if (a[i] > 1) c[a[i] - 1] = (2 * c[a[i] - 1] + b[a[i] - 2]) % M;
    c[a[i] + 1] = (2 * c[a[i] + 1]) % M;
  }
  for (i = 0; i < n + 1; i++) {
    ans = (ans + b[i] + c[i]) % M;
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  sub();
  auto start1 = high_resolution_clock::now();
  long long t = 1, s;
  cin >> t;
  for (s = 1; s < t + 1; s++) {
    solve();
  }
  auto stop1 = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop1 - start1);
  cerr << "Time: " << (double)duration.count() / 1000 << " ms" << endl;
  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;
using namespace chrono;
template <class T>
void show(vector<vector<T>> &a) {
  for (auto &i : a) {
    for (T &j : i) cout << j << ' ';
    cout << endl;
  }
}
template <class T>
void show(vector<T> &a) {
  for (T i : a) cout << i << ' ';
  cout << endl;
}
template <class T>
void show(vector<pair<T, T>> &a) {
  for (auto &i : a) {
    cout << i.first << ' ' << i.second << endl;
  }
}
template <class T>
void show(pair<T, T> p) {
  cout << p.first << ' ' << p.second << endl;
}
void show(vector<string> &a) {
  for (auto &i : a) cout << i << endl;
}
template <class T>
void take(vector<T> &a) {
  for (auto &i : a) cin >> i;
}
long long digits(long long n) {
  long long k = 0;
  while (n) k++, n /= 10;
  return k;
}
bool isprime(long long n) {
  long long i;
  if (n == 1) return 0;
  if (n == 2) return 1;
  if (n % 2 == 0) return 0;
  for (i = 3; i * i <= n; i += 2) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long power(long long x, long long y, long long MOD) {
  long long ans = 1;
  x %= MOD;
  while (y > 0) {
    if (y % 2 == 1) ans = (ans * x) % MOD;
    y /= 2;
    x = (x * x) % MOD;
  }
  return ans;
}
void sieve(vector<bool> &a, long long n) {
  long long i, j;
  a[1] = 0;
  a[0] = 0;
  long long h = sqrt(n + 1);
  for (i = 2; i < h; i++) {
    if (!a[i]) continue;
    for (j = 2 * i; j < n + 1; j += i) a[j] = 0;
  }
}
void load(unordered_map<long long, long long> &g) { g.max_load_factor(0.25); }
vector<long long> bits(long long n) {
  vector<long long> a;
  long long i;
  for (i = 0; i < 33; i++) {
    a.push_back(n % 2);
    n /= 2;
  }
  reverse(a.begin(), a.end());
  return a;
}
void sub() {}
const long long N = 1e9 + 7;
const long long M = 998244353;
void solve() {
  long long i, j, n, m;
  cin >> n;
  vector<long long> a(n);
  take(a);
  long long ans = 0;
  for (i = 0; i < n; i++)
    if (a[i] == 1) ans++;
  ans = (power(2, ans, M) - 1 + M) % M;
  vector<long long> b(n + 3), c(n + 3);
  for (i = 0; i < n; i++) {
    if (a[i] == 0) b[0] = (2 * b[0] + 1) % M;
    if (a[i] == 1) b[1] = (2 * b[1] + b[0]) % M;
    if (a[i] > 1) b[a[i]] = (2 * b[a[i]] + b[a[i] - 1]) % M;
    if (a[i] > 1) c[a[i] - 1] = (2 * c[a[i] - 1] + b[a[i] - 2]) % M;
    c[a[i] + 1] = (2 * c[a[i] + 1]) % M;
  }
  for (i = 0; i < n + 1; i++) {
    ans = (ans + b[i] + c[i]) % M;
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  sub();
  auto start1 = high_resolution_clock::now();
  long long t = 1, s;
  cin >> t;
  for (s = 1; s < t + 1; s++) {
    solve();
  }
  auto stop1 = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop1 - start1);
  cerr << "Time: " << (double)duration.count() / 1000 << " ms" << endl;
  return 0;
}

```