#include <bits/stdc++.h>
using namespace std;
const int MAXL = 1 << 22;
char i_str[MAXL], o_str[MAXL], *i_s, *i_t;
int o_t;
inline char gc() {
  if (i_s == i_t) {
    i_s = i_str;
    i_t = i_s + fread(i_str, 1, MAXL, stdin);
    return i_s == i_t ? EOF : *i_s++;
  } else
    return *i_s++;
}
inline int read() {
  int x = 0, f = 0;
  char ch = gc();
  for (; ch < '0' || ch > '9'; ch = gc())
    if (ch == '-') f = 1;
  for (; ch >= '0' && ch <= '9'; ch = gc()) x = x * 10 + (ch ^ 48);
  return f ? -x : x;
}
inline void pc(char x) {
  o_str[o_t++] = x;
  if (o_t == MAXL) fwrite(o_str, 1, o_t, stdout), o_t = 0;
}
void write(int x) {
  if (x < 0) pc('-'), x = -x;
  if (x > 9) write(x / 10);
  pc(x % 10 ^ 48);
}
const int N = 5e5 + 5, mod = 998244353;
int D, f[N][3];
inline void Solve() {
  int n = read(), ans = 0;
  for (register int i = 0; i <= n + 2; ++i) f[i][0] = f[i][1] = f[i][2] = 0;
  for (register int i = 1; i <= n; ++i) {
    int x = read();
    f[x + 2][1] = (f[x + 2][1] + f[x + 2][1]) % mod;
    if (!x) {
      f[x][0] = (f[x][0] + f[x][0] + 1) % mod;
      continue;
    }
    f[x][0] = ((f[x][0] + f[x][0]) % mod + f[x - 1][0]) % mod;
    if (x == 1) {
      f[x][1] = (f[x][1] + f[x][1] + 1) % mod;
      continue;
    }
    f[x][1] = ((f[x][1] + f[x][1]) % mod + f[x - 2][0]) % mod;
  }
  for (register int i = 0; i <= n + 2; ++i)
    ans = (((ans + f[i][0]) % mod + f[i][1]) % mod + f[i][2]) % mod;
  write(ans), pc('\n');
  return;
}
signed main() {
  int T = read();
  while (T--) Solve();
  fwrite(o_str, 1, o_t, stdout), o_t = 0;
  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 MAXL = 1 << 22;
char i_str[MAXL], o_str[MAXL], *i_s, *i_t;
int o_t;
inline char gc() {
  if (i_s == i_t) {
    i_s = i_str;
    i_t = i_s + fread(i_str, 1, MAXL, stdin);
    return i_s == i_t ? EOF : *i_s++;
  } else
    return *i_s++;
}
inline int read() {
  int x = 0, f = 0;
  char ch = gc();
  for (; ch < '0' || ch > '9'; ch = gc())
    if (ch == '-') f = 1;
  for (; ch >= '0' && ch <= '9'; ch = gc()) x = x * 10 + (ch ^ 48);
  return f ? -x : x;
}
inline void pc(char x) {
  o_str[o_t++] = x;
  if (o_t == MAXL) fwrite(o_str, 1, o_t, stdout), o_t = 0;
}
void write(int x) {
  if (x < 0) pc('-'), x = -x;
  if (x > 9) write(x / 10);
  pc(x % 10 ^ 48);
}
const int N = 5e5 + 5, mod = 998244353;
int D, f[N][3];
inline void Solve() {
  int n = read(), ans = 0;
  for (register int i = 0; i <= n + 2; ++i) f[i][0] = f[i][1] = f[i][2] = 0;
  for (register int i = 1; i <= n; ++i) {
    int x = read();
    f[x + 2][1] = (f[x + 2][1] + f[x + 2][1]) % mod;
    if (!x) {
      f[x][0] = (f[x][0] + f[x][0] + 1) % mod;
      continue;
    }
    f[x][0] = ((f[x][0] + f[x][0]) % mod + f[x - 1][0]) % mod;
    if (x == 1) {
      f[x][1] = (f[x][1] + f[x][1] + 1) % mod;
      continue;
    }
    f[x][1] = ((f[x][1] + f[x][1]) % mod + f[x - 2][0]) % mod;
  }
  for (register int i = 0; i <= n + 2; ++i)
    ans = (((ans + f[i][0]) % mod + f[i][1]) % mod + f[i][2]) % mod;
  write(ans), pc('\n');
  return;
}
signed main() {
  int T = read();
  while (T--) Solve();
  fwrite(o_str, 1, o_t, stdout), o_t = 0;
  return 0;
}

```