#include <bits/stdc++.h>
using namespace std;
const long long N = 5e5 + 5;
const long long M = 5e3 + 5;
const long long mod = 998244353;
long long n;
long long a[N];
long long dp1[N];
long long dp2[N];
long long add(long long t1, long long t2) {
  return ((t1 % mod) + (t2 % mod)) % mod;
}
long long sub(long long t1, long long t2) {
  long long res = ((t1 % mod) - (t2 % mod)) % mod;
  while (res < 0) res += mod;
  res %= mod;
  return res;
}
int main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    dp1[0] = dp2[0] = dp1[n + 1] = dp2[n + 1] = 0;
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      dp1[i] = 0;
      dp2[i] = 0;
    }
    dp1[0] = 1;
    for (long long i = 1; i <= n; i++) {
      long long x = a[i];
      dp1[x + 1] = add(dp1[x + 1], dp1[x + 1]);
      dp1[x + 1] = add(dp1[x + 1], dp1[x]);
      if (x > 0) dp2[x - 1] = add(dp2[x - 1], dp2[x - 1]);
      if (x > 0) dp2[x - 1] = add(dp2[x - 1], dp1[x - 1]);
      dp2[x + 1] = add(dp2[x + 1], dp2[x + 1]);
    }
    long long ans = 0;
    for (long long i = 0; i <= n + 1; i++) {
      ans = add(ans, dp1[i]);
      ans = add(ans, dp2[i]);
    }
    printf("%lld\n", sub(ans, 1ll));
  }
}

### Prompt

Please create a solution in cpp 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 long long N = 5e5 + 5;
const long long M = 5e3 + 5;
const long long mod = 998244353;
long long n;
long long a[N];
long long dp1[N];
long long dp2[N];
long long add(long long t1, long long t2) {
  return ((t1 % mod) + (t2 % mod)) % mod;
}
long long sub(long long t1, long long t2) {
  long long res = ((t1 % mod) - (t2 % mod)) % mod;
  while (res < 0) res += mod;
  res %= mod;
  return res;
}
int main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    dp1[0] = dp2[0] = dp1[n + 1] = dp2[n + 1] = 0;
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      dp1[i] = 0;
      dp2[i] = 0;
    }
    dp1[0] = 1;
    for (long long i = 1; i <= n; i++) {
      long long x = a[i];
      dp1[x + 1] = add(dp1[x + 1], dp1[x + 1]);
      dp1[x + 1] = add(dp1[x + 1], dp1[x]);
      if (x > 0) dp2[x - 1] = add(dp2[x - 1], dp2[x - 1]);
      if (x > 0) dp2[x - 1] = add(dp2[x - 1], dp1[x - 1]);
      dp2[x + 1] = add(dp2[x + 1], dp2[x + 1]);
    }
    long long ans = 0;
    for (long long i = 0; i <= n + 1; i++) {
      ans = add(ans, dp1[i]);
      ans = add(ans, dp2[i]);
    }
    printf("%lld\n", sub(ans, 1ll));
  }
}

```