#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return a % b ? gcd(b, a % b) : b; }
const long long mod = 998244353;
const long long inf = 1e18;
long long qpow(long long a, long long b) {
  long long t = 1;
  while (b != 0) {
    if (b & 1) t = (t * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return t;
}
template <typename T>
void re(T& a) {
  for (auto& t : a) cin >> t;
}
template <typename T>
void pr(T& a) {
  for (auto& t : a) cout << t << " ";
  cout << endl;
}
const int N = 5e5 + 5;
long long a[N];
long long dp[N];
long long pd[N];
int n;
void add(long long& x, long long y) { x = ((x + y) % mod + mod) % mod; }
void solve() {
  cin >> n;
  for (int i = 0; i <= n; ++i) {
    dp[i] = 0;
    pd[i] = 0;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    long long x = a[i];
    long long tmp = dp[x];
    if (x == 0) {
      add(tmp, 1);
    } else {
      add(tmp, dp[x - 1]);
    }
    add(dp[x], tmp);
    x += 2;
    add(pd[x], pd[x]);
    x -= 2;
    if (x == 0) {
      continue;
    }
    tmp = pd[x];
    if (x == 1) {
      add(tmp, 1);
    } else {
      add(tmp, dp[x - 2]);
    }
    add(pd[x], tmp);
  }
  long long ans = 0;
  for (int i = 0; i <= n; ++i) {
    add(ans, dp[i]);
    add(ans, pd[i]);
  }
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 gcd(long long a, long long b) { return a % b ? gcd(b, a % b) : b; }
const long long mod = 998244353;
const long long inf = 1e18;
long long qpow(long long a, long long b) {
  long long t = 1;
  while (b != 0) {
    if (b & 1) t = (t * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return t;
}
template <typename T>
void re(T& a) {
  for (auto& t : a) cin >> t;
}
template <typename T>
void pr(T& a) {
  for (auto& t : a) cout << t << " ";
  cout << endl;
}
const int N = 5e5 + 5;
long long a[N];
long long dp[N];
long long pd[N];
int n;
void add(long long& x, long long y) { x = ((x + y) % mod + mod) % mod; }
void solve() {
  cin >> n;
  for (int i = 0; i <= n; ++i) {
    dp[i] = 0;
    pd[i] = 0;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    long long x = a[i];
    long long tmp = dp[x];
    if (x == 0) {
      add(tmp, 1);
    } else {
      add(tmp, dp[x - 1]);
    }
    add(dp[x], tmp);
    x += 2;
    add(pd[x], pd[x]);
    x -= 2;
    if (x == 0) {
      continue;
    }
    tmp = pd[x];
    if (x == 1) {
      add(tmp, 1);
    } else {
      add(tmp, dp[x - 2]);
    }
    add(pd[x], tmp);
  }
  long long ans = 0;
  for (int i = 0; i <= n; ++i) {
    add(ans, dp[i]);
    add(ans, pd[i]);
  }
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```