#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9;
const long long INFF = 1e18;
const long long MAXN = 510;
const long long MOD = 998244353;
const double PI = acos(-1.0);
const double INFD = 1E9;
const double EPS = 1e-9;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long getTC() {
  if (1) {
    long long t;
    cin >> t;
    return t;
  } else
    return 1;
}
long long add(long long x, long long y) {
  return ((x % MOD) + (y % MOD)) % MOD;
}
long long sub(long long x, long long y) {
  long long ret = ((x % MOD) - (y % MOD)) % MOD;
  if (ret < 0) ret += MOD;
  return ret;
}
long long mul(long long x, long long y) {
  return ((x % MOD) * (y % MOD)) % MOD;
}
long long modpow(long long x, long long y) {
  if (!y) return 1;
  long long z = modpow(x, y / 2);
  z = mul(z, z);
  if (y & 1) z = mul(z, x);
  return z;
}
long long reverse(long long x) { return modpow(x, MOD - 2); }
long long bagi(long long x, long long y) { return mul(x, reverse(y)); }
void solve() {
  long long n;
  cin >> n;
  vector<long long> tab(n);
  for (auto &it : tab) cin >> it;
  vector<vector<long long>> dp(2, vector<long long>(n + 5, 0));
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    if (tab[i] == 0) {
      dp[0][0] = add(dp[0][0], dp[0][0]);
      dp[0][0] = add(dp[0][0], 1);
    } else {
      dp[1][tab[i]] = add(dp[1][tab[i]], dp[1][tab[i]]);
      dp[0][tab[i]] = add(dp[0][tab[i]], dp[0][tab[i]]);
      if (tab[i] >= 2) dp[1][tab[i]] = add(dp[1][tab[i]], dp[0][tab[i] - 2]);
      dp[0][tab[i]] = add(dp[0][tab[i]], dp[0][tab[i] - 1]);
      if (tab[i] == 1) dp[1][tab[i]] = add(dp[1][tab[i]], 1);
    }
    dp[1][tab[i] + 2] = add(dp[1][tab[i] + 2], dp[1][tab[i] + 2]);
  }
  for (long long i = 0; i <= n; i++) {
    ans = add(ans, dp[0][i]);
    ans = add(ans, dp[1][i]);
  }
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = getTC();
  while (t--) {
    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 INF = 1e9;
const long long INFF = 1e18;
const long long MAXN = 510;
const long long MOD = 998244353;
const double PI = acos(-1.0);
const double INFD = 1E9;
const double EPS = 1e-9;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long getTC() {
  if (1) {
    long long t;
    cin >> t;
    return t;
  } else
    return 1;
}
long long add(long long x, long long y) {
  return ((x % MOD) + (y % MOD)) % MOD;
}
long long sub(long long x, long long y) {
  long long ret = ((x % MOD) - (y % MOD)) % MOD;
  if (ret < 0) ret += MOD;
  return ret;
}
long long mul(long long x, long long y) {
  return ((x % MOD) * (y % MOD)) % MOD;
}
long long modpow(long long x, long long y) {
  if (!y) return 1;
  long long z = modpow(x, y / 2);
  z = mul(z, z);
  if (y & 1) z = mul(z, x);
  return z;
}
long long reverse(long long x) { return modpow(x, MOD - 2); }
long long bagi(long long x, long long y) { return mul(x, reverse(y)); }
void solve() {
  long long n;
  cin >> n;
  vector<long long> tab(n);
  for (auto &it : tab) cin >> it;
  vector<vector<long long>> dp(2, vector<long long>(n + 5, 0));
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    if (tab[i] == 0) {
      dp[0][0] = add(dp[0][0], dp[0][0]);
      dp[0][0] = add(dp[0][0], 1);
    } else {
      dp[1][tab[i]] = add(dp[1][tab[i]], dp[1][tab[i]]);
      dp[0][tab[i]] = add(dp[0][tab[i]], dp[0][tab[i]]);
      if (tab[i] >= 2) dp[1][tab[i]] = add(dp[1][tab[i]], dp[0][tab[i] - 2]);
      dp[0][tab[i]] = add(dp[0][tab[i]], dp[0][tab[i] - 1]);
      if (tab[i] == 1) dp[1][tab[i]] = add(dp[1][tab[i]], 1);
    }
    dp[1][tab[i] + 2] = add(dp[1][tab[i] + 2], dp[1][tab[i] + 2]);
  }
  for (long long i = 0; i <= n; i++) {
    ans = add(ans, dp[0][i]);
    ans = add(ans, dp[1][i]);
  }
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = getTC();
  while (t--) {
    solve();
  }
  return 0;
}

```