#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long binp(long long a, long long p) {
  if (p == 0) return 1;
  if (p & 1) {
    return ((long long)a * binp(a, p - 1)) % mod;
  } else {
    long long t = binp(a, p / 2);
    return (t * t) % mod;
  }
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<vector<long long>> pos(n + 1);
  for (long long i = 0; i < n; i++) pos[a[i]].push_back(i);
  vector<long long> dpprev(n, 0);
  vector<long long> dpcur(n, 0);
  vector<long long> dpans(n, 0);
  long long cnt0 = 0;
  long long res = 0;
  long long cnt1 = 0;
  for (long long i = 0; i < n; i++)
    if (a[i] == 1) cnt1++;
  res = binp(2, cnt1) - 1;
  long long cnt02 = 0;
  for (long long i = 0; i < n; i++)
    if (a[i] == 0 || a[i] == 2) cnt02++;
  for (long long i = 0; i < n; i++) {
    if (a[i] != 0 && a[i] != 2) continue;
    if (a[i] == 0) res = (res + binp(2, cnt02 - 1)) % mod;
    cnt02--;
  }
  for (long long i = 0; i < n; i++) {
    if (a[i]) continue;
    dpprev[i] = binp(2, cnt0);
    cnt0++;
  }
  for (long long i = 1; i <= n; i++) {
    long long id = 0;
    long long idn = 0;
    for (auto e : pos[i]) dpcur[e] = dpans[e] = 0;
    if (pos[i].size()) {
      while (id < pos[i - 1].size() && pos[i - 1][id] < pos[i][0]) {
        dpcur[pos[i][0]] = (dpcur[pos[i][0]] + dpprev[pos[i - 1][id]]) % mod;
        dpans[pos[i][0]] = (dpans[pos[i][0]] + dpprev[pos[i - 1][id]]) % mod;
        id++;
      }
      long long cntcurnext = 0;
      if (i + 2 <= n)
        cntcurnext = pos[i].size() + pos[i + 2].size();
      else
        cntcurnext = pos[i].size();
      while (i + 2 <= n && idn < pos[i + 2].size() &&
             pos[i + 2][idn] < pos[i][0]) {
        cntcurnext--;
        idn++;
      }
      res = (res +
             ((long long)dpcur[pos[i][0]] * binp(2, cntcurnext - 1)) % mod) %
            mod;
      cntcurnext--;
      for (long long j = 1; j < pos[i].size(); j++) {
        dpans[pos[i][j]] = dpans[pos[i][j - 1]];
        dpcur[pos[i][j]] = (2 * dpcur[pos[i][j - 1]]) % mod;
        while (i + 2 <= n && idn < pos[i + 2].size() &&
               pos[i + 2][idn] < pos[i][j]) {
          cntcurnext--;
          idn++;
        }
        while (id < pos[i - 1].size() && pos[i - 1][id] < pos[i][j]) {
          dpcur[pos[i][j]] = (dpcur[pos[i][j]] + dpprev[pos[i - 1][id]]) % mod;
          dpans[pos[i][j]] = (dpans[pos[i][j]] + dpprev[pos[i - 1][id]]) % mod;
          id++;
        }
        res = (res +
               ((long long)dpans[pos[i][j]] * binp(2, cntcurnext - 1)) % mod) %
              mod;
        cntcurnext--;
      }
    }
    for (auto e : pos[i]) dpprev[e] = dpcur[e];
  }
  cout << res << "\n";
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test_cnt = 1;
  cin >> test_cnt;
  for (long long test = 1; test <= test_cnt; test++) {
    solve();
  }
  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;
const long long mod = 998244353;
long long binp(long long a, long long p) {
  if (p == 0) return 1;
  if (p & 1) {
    return ((long long)a * binp(a, p - 1)) % mod;
  } else {
    long long t = binp(a, p / 2);
    return (t * t) % mod;
  }
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<vector<long long>> pos(n + 1);
  for (long long i = 0; i < n; i++) pos[a[i]].push_back(i);
  vector<long long> dpprev(n, 0);
  vector<long long> dpcur(n, 0);
  vector<long long> dpans(n, 0);
  long long cnt0 = 0;
  long long res = 0;
  long long cnt1 = 0;
  for (long long i = 0; i < n; i++)
    if (a[i] == 1) cnt1++;
  res = binp(2, cnt1) - 1;
  long long cnt02 = 0;
  for (long long i = 0; i < n; i++)
    if (a[i] == 0 || a[i] == 2) cnt02++;
  for (long long i = 0; i < n; i++) {
    if (a[i] != 0 && a[i] != 2) continue;
    if (a[i] == 0) res = (res + binp(2, cnt02 - 1)) % mod;
    cnt02--;
  }
  for (long long i = 0; i < n; i++) {
    if (a[i]) continue;
    dpprev[i] = binp(2, cnt0);
    cnt0++;
  }
  for (long long i = 1; i <= n; i++) {
    long long id = 0;
    long long idn = 0;
    for (auto e : pos[i]) dpcur[e] = dpans[e] = 0;
    if (pos[i].size()) {
      while (id < pos[i - 1].size() && pos[i - 1][id] < pos[i][0]) {
        dpcur[pos[i][0]] = (dpcur[pos[i][0]] + dpprev[pos[i - 1][id]]) % mod;
        dpans[pos[i][0]] = (dpans[pos[i][0]] + dpprev[pos[i - 1][id]]) % mod;
        id++;
      }
      long long cntcurnext = 0;
      if (i + 2 <= n)
        cntcurnext = pos[i].size() + pos[i + 2].size();
      else
        cntcurnext = pos[i].size();
      while (i + 2 <= n && idn < pos[i + 2].size() &&
             pos[i + 2][idn] < pos[i][0]) {
        cntcurnext--;
        idn++;
      }
      res = (res +
             ((long long)dpcur[pos[i][0]] * binp(2, cntcurnext - 1)) % mod) %
            mod;
      cntcurnext--;
      for (long long j = 1; j < pos[i].size(); j++) {
        dpans[pos[i][j]] = dpans[pos[i][j - 1]];
        dpcur[pos[i][j]] = (2 * dpcur[pos[i][j - 1]]) % mod;
        while (i + 2 <= n && idn < pos[i + 2].size() &&
               pos[i + 2][idn] < pos[i][j]) {
          cntcurnext--;
          idn++;
        }
        while (id < pos[i - 1].size() && pos[i - 1][id] < pos[i][j]) {
          dpcur[pos[i][j]] = (dpcur[pos[i][j]] + dpprev[pos[i - 1][id]]) % mod;
          dpans[pos[i][j]] = (dpans[pos[i][j]] + dpprev[pos[i - 1][id]]) % mod;
          id++;
        }
        res = (res +
               ((long long)dpans[pos[i][j]] * binp(2, cntcurnext - 1)) % mod) %
              mod;
        cntcurnext--;
      }
    }
    for (auto e : pos[i]) dpprev[e] = dpcur[e];
  }
  cout << res << "\n";
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test_cnt = 1;
  cin >> test_cnt;
  for (long long test = 1; test <= test_cnt; test++) {
    solve();
  }
  return 0;
}

```