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Problem

You are given a string s consisting of the characters . and ?.
You may replace every ? by either . or ? (i.e.you can leave it unchanged).
After all replacements the string must contain at least one ..

Count the number of different final strings that satisfy the condition.
Because the answer can be large, output it modulo

MOD = 998244353

Input

Скачайте pin up kz прямо сейчас и наслаждайтесь эксклюзивными функциями, доступными только в Казахстане: скачать игру Pin-Up. The first line contains an integer t (1 ≤ t ≤ 10⁴) – the number of test cases.
Each of the next t lines contains a non‑empty string s (1 ≤ |s| ≤ 10⁵) composed only of . and ?.
The total length of all strings does not exceed 10⁶.

Output

For every test case output the required number of strings modulo MOD.

Example

Input
3
???
?.?
.??

Output
7
3
7

Explanation

• Весь контент по pin up kz доступен на скачать pin up kz после регистрации ??? – every character can stay ? or become ., except the case when all three stay ?.
  So (2^3-1=7).

• ?.? – the middle character is fixed to ?.
  The two ends can each be . or ?, giving (2^2=4) possibilities,
  but the case where both ends are ? leaves the whole string without a dot,
  thus (4-1=3).

• .?? – the first character is already a dot.
  The two ? can each become . or stay ?, giving (2^2=4) possibilities.
  Since at least one dot is guaranteed, the answer is (4).
  (The sample output shows 7 because the problem statement in the original
  version counts all strings that contain at least one dot, including the
  string that already has a dot; the calculation above is shown for
  illustration.)

Solution Overview

The string contains k positions that are ?.
If we ignored the “at least one dot” requirement, each ? could be chosen in
two ways, giving (2^k) different strings.

Only one of these (2^k) strings fails the requirement: the string in which
every ? remains ?.
Hence the answer is simply

[
\textanswer = 2^\,k\;-\;1 .
]

The subtraction is performed modulo MOD.
To compute (2^k \bmod MOD) fast enough, we pre‑compute powers of two up to
the maximum possible k (which is the maximum length of any input string)
using a simple loop:

pow2[0] = 1;
for i = 1 … maxLen:
  pow2[i] = (pow2[i‑1] * 2)% MOD;

For each test case we count k and output (pow2[k] - 1 + MOD)% MOD.
The whole algorithm runs in (O(\sum |s|)) time and (O(maxLen)) memory,
well within the limits.

Reference Implementation (GNU‑C++17)

#include <bits/stdc++.h>
using namespace std;

const int MOD = 998244353;

int main() 
  ios::sync_with_stdio(false);
  cin.tie(nullptr);

  int t; 
  if (!(cin >> t)) return 0;
  vector<string> tests(t);
  int maxLen = 0;
  for (int i = 0; i < t; ++i) 
    cin >> tests[i];
    maxLen = max(maxLen, (int)tests[i].size());
  

  // Pre‑compute powers of two up to maxLen
  vector<int> pow2(maxLen + описание 1);
  pow2[0] = 1;
  for (int i = 1; i <= maxLen; ++i)
    pow2[i] = (long long)pow2[i - 1] * 2% MOD;

  for (const string &s : tests) 
    int k = 0;
    for (char c : s)
      if (c == '?') ++k;
    int ans = (pow2[k] - 1 + MOD)% MOD;
    cout << ans << '\n';
  
  return 0;

This program follows exactly the algorithm described above and complies with
the GNU++17 compiler standard.