#include <bits/stdc++.h>
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return x * f;
}
const long long N = 5e5 + 5, mod = 998244353;
long long a[N], dp[N][3], t[N][3], mi[N], n, ans, res;
signed main(void) {
  mi[0] = 1;
  for (long long i = 1; i <= 500000; i++) mi[i] = mi[i - 1] * 2, mi[i] %= mod;
  long long T = read();
  while (T--) {
    n = read(), ans = res = 0;
    for (long long i = 1; i <= n; i++) a[i] = read();
    for (long long i = 1; i <= n; i++) {
      if (a[i] >= 1) dp[i][0] += t[a[i] - 1][0];
      dp[i][0] += t[a[i]][0], dp[i][0] %= mod;
      if (a[i] >= 2) dp[i][1] += t[a[i] - 2][0] + t[a[i] - 2][2];
      dp[i][1] += t[a[i]][1], dp[i][1] %= mod;
      dp[i][2] += t[a[i] + 2][1], dp[i][2] += t[a[i]][2], dp[i][2] %= mod;
      if (a[i] == 1) ans += mi[res], res++, ans %= mod;
      if (a[i] == 0) dp[i][0] += 1;
      t[a[i]][0] += dp[i][0], t[a[i]][1] += dp[i][1], t[a[i]][0] %= mod,
          t[a[i]][1] %= mod;
      t[a[i]][2] += dp[i][2], t[a[i]][2] %= mod;
    }
    for (long long i = 1; i <= n; i++) {
      ans += dp[i][0] + dp[i][1] + dp[i][2], ans %= mod;
      t[a[i]][0] = t[a[i]][1] = t[a[i]][2] = 0,
      dp[i][0] = dp[i][1] = dp[i][2] = 0;
    }
    printf("%lld\n", ans);
  }
  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>
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return x * f;
}
const long long N = 5e5 + 5, mod = 998244353;
long long a[N], dp[N][3], t[N][3], mi[N], n, ans, res;
signed main(void) {
  mi[0] = 1;
  for (long long i = 1; i <= 500000; i++) mi[i] = mi[i - 1] * 2, mi[i] %= mod;
  long long T = read();
  while (T--) {
    n = read(), ans = res = 0;
    for (long long i = 1; i <= n; i++) a[i] = read();
    for (long long i = 1; i <= n; i++) {
      if (a[i] >= 1) dp[i][0] += t[a[i] - 1][0];
      dp[i][0] += t[a[i]][0], dp[i][0] %= mod;
      if (a[i] >= 2) dp[i][1] += t[a[i] - 2][0] + t[a[i] - 2][2];
      dp[i][1] += t[a[i]][1], dp[i][1] %= mod;
      dp[i][2] += t[a[i] + 2][1], dp[i][2] += t[a[i]][2], dp[i][2] %= mod;
      if (a[i] == 1) ans += mi[res], res++, ans %= mod;
      if (a[i] == 0) dp[i][0] += 1;
      t[a[i]][0] += dp[i][0], t[a[i]][1] += dp[i][1], t[a[i]][0] %= mod,
          t[a[i]][1] %= mod;
      t[a[i]][2] += dp[i][2], t[a[i]][2] %= mod;
    }
    for (long long i = 1; i <= n; i++) {
      ans += dp[i][0] + dp[i][1] + dp[i][2], ans %= mod;
      t[a[i]][0] = t[a[i]][1] = t[a[i]][2] = 0,
      dp[i][0] = dp[i][1] = dp[i][2] = 0;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```