description: A binary string is a string that consists of characters 0 and 1. A bi-table is a table that has exactly two rows of equal length, each being a binary string.
  
  Let \operatorname{MEX} of a bi-table be the smallest digit among 0, 1, or 2 that does not occur in the bi-table. For example, \operatorname{MEX} for \begin{bmatrix} 0011\\\ 1010 \end{bmatrix} is 2, because 0 and 1 occur in the bi-table at least once. \operatorname{MEX} for \begin{bmatrix} 111\\\ 111 \end{bmatrix} is 0, because 0 and 2 do not occur in the bi-table, and 0 < 2.
  
  You are given a bi-table with n columns. You should cut it into any number of bi-tables (each consisting of consecutive columns) so that each column is in exactly one bi-table. It is possible to cut the bi-table into a single bi-table — the whole bi-table.
  
  What is the maximal sum of \operatorname{MEX} of all resulting bi-tables can be?
  
  Input
  
  The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
  
  The first line of the description of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of columns in the bi-table.
  
  Each of the next two lines contains a binary string of length n — the rows of the bi-table.
  
  It's guaranteed that the sum of n over all test cases does not exceed 10^5.
  
  Output
  
  For each test case print a single integer — the maximal sum of \operatorname{MEX} of all bi-tables that it is possible to get by cutting the given bi-table optimally.
  
  Example
  
  Input
  
  
  4
  7
  0101000
  1101100
  5
  01100
  10101
  2
  01
  01
  6
  000000
  111111
  
  
  Output
  
  
  8
  8
  2
  12
  
  Note
  
  In the first test case you can cut the bi-table as follows:
  
    * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
    * \begin{bmatrix} 10\\\ 10 \end{bmatrix}, its \operatorname{MEX} is 2.
    * \begin{bmatrix} 1\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 0.
    * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
    * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.
    * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.
  
  
  
  The sum of \operatorname{MEX} is 8.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int mod = 1e9 + 7;
  int t, n;
  long long cnt, ans;
  string s, c;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t--) {
      cin >> n >> s >> c;
      n = s.size();
      ans = cnt = 0;
      bool flag = 0, ff1 = 1, ff2 = 1, f0 = 1, f00 = 1;
      for (int i = 0; i < n; i++) {
        if (s[i] != c[i])
          ans += 2, flag = 0;
        else if (s[i] == '1' && c[i] == '1') {
          flag = 1;
        } else {
          if (flag)
            ans += 2, flag = 0;
          else {
            if (i != n - 1 && s[i + 1] == '1' && c[i + 1] == '1') {
              ans += 2;
              i++;
            } else
              ans++;
          }
        }
      }
      flag = 0;
      for (int i = n - 1; i >= 0; i--) {
        if (s[i] != c[i])
          cnt += 2, flag = 0;
        else if (s[i] == '1' && c[i] == '1') {
          flag = 1;
        } else {
          if (flag)
            cnt += 2, flag = 0;
          else {
            if (i && s[i - 1] == '1' && c[i - 1] == '1') {
              cnt += 2;
              i--;
            } else
              cnt++;
          }
        }
      }
      cout << max(ans, cnt) << endl;
    }
    return 0;
  }
  