#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10;
const long long INF = 1e9 + 10;
long long MOD = 998244353;
long double EPS = 1e-12;
void fastIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long t, n, a[MAXN];
long long dp1[MAXN], dp2[MAXN], cnt[MAXN];
int sum(int x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
  return x;
}
int mul(int x, int y) { return x * 1ll * y % MOD; }
int bin_pow(int x, int n) {
  int res = 1;
  while (n > 0) {
    if (n & 1) res = mul(res, x);
    x = mul(x, x);
    n >>= 1;
  }
  return res;
}
int main() {
  fastIO();
  cin >> t;
  while (t--) {
    cin >> n;
    fill(cnt, cnt + n + 1, 0);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      cnt[a[i]]++;
    }
    fill(dp1, dp1 + n + 10, 0);
    fill(dp2, dp2 + n + 10, 0);
    int ans = 0;
    for (int i = 0; i < n; i++) {
      int val = a[i];
      cnt[val]--;
      dp1[val] = sum(dp1[val], dp1[val]);
      if (val > 0) {
        dp1[val] = sum(dp1[val], dp1[val - 1]);
      } else {
        dp1[val] = sum(dp1[val], 1);
      }
      if (val > 1) {
        ans = sum(ans, mul(dp1[val - 2], bin_pow(2, cnt[val - 2] + cnt[val])));
      }
      if (val == 1) {
        ans = sum(ans, bin_pow(2, cnt[val]));
      }
    }
    for (int i = 0; i <= n; i++) {
      ans = sum(ans, dp1[i]);
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to 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 int MAXN = 1e6 + 10;
const long long INF = 1e9 + 10;
long long MOD = 998244353;
long double EPS = 1e-12;
void fastIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long t, n, a[MAXN];
long long dp1[MAXN], dp2[MAXN], cnt[MAXN];
int sum(int x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
  return x;
}
int mul(int x, int y) { return x * 1ll * y % MOD; }
int bin_pow(int x, int n) {
  int res = 1;
  while (n > 0) {
    if (n & 1) res = mul(res, x);
    x = mul(x, x);
    n >>= 1;
  }
  return res;
}
int main() {
  fastIO();
  cin >> t;
  while (t--) {
    cin >> n;
    fill(cnt, cnt + n + 1, 0);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      cnt[a[i]]++;
    }
    fill(dp1, dp1 + n + 10, 0);
    fill(dp2, dp2 + n + 10, 0);
    int ans = 0;
    for (int i = 0; i < n; i++) {
      int val = a[i];
      cnt[val]--;
      dp1[val] = sum(dp1[val], dp1[val]);
      if (val > 0) {
        dp1[val] = sum(dp1[val], dp1[val - 1]);
      } else {
        dp1[val] = sum(dp1[val], 1);
      }
      if (val > 1) {
        ans = sum(ans, mul(dp1[val - 2], bin_pow(2, cnt[val - 2] + cnt[val])));
      }
      if (val == 1) {
        ans = sum(ans, bin_pow(2, cnt[val]));
      }
    }
    for (int i = 0; i <= n; i++) {
      ans = sum(ans, dp1[i]);
    }
    cout << ans << '\n';
  }
  return 0;
}

```