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;
  void solve() {
    long long int n;
    cin >> n;
    string s1, s2;
    cin >> s1;
    cin >> s2;
    if (count(s1.begin(), s1.end(), '0') == n &&
        count(s2.begin(), s2.end(), '0') == n) {
      cout << n << "\n";
      return;
    } else if (count(s1.begin(), s1.end(), '1') == n &&
               count(s2.begin(), s2.end(), '1') == n) {
      cout << 0 << "\n";
      return;
    } else if (count(s1.begin(), s1.end(), '1') == n &&
               count(s2.begin(), s2.end(), '0') == n) {
      cout << 2 * n << "\n";
      return;
    } else if (count(s1.begin(), s1.end(), '0') == n &&
               count(s2.begin(), s2.end(), '1') == n) {
      cout << 2 * n << "\n";
      return;
    }
    long long int mex = 0;
    string aa = "";
    vector<string> a;
    for (long long int i = 0; i < n; i++) {
      if (s1.at(i) != s2.at(i)) {
        mex += 2;
        if (aa != "" || count(aa.begin(), aa.end(), '1') != aa.size()) {
          a.push_back(aa);
          aa = "";
        }
      } else {
        aa = aa + s1.at(i);
      }
    }
    if (aa != "" || count(aa.begin(), aa.end(), '1') != aa.size()) {
      a.push_back(aa);
      aa = "";
    }
    vector<string>::iterator i;
    for (i = a.begin(); i != a.end(); i++) {
      string s = *i;
      if (count(s.begin(), s.end(), '0') == s.size()) {
        mex += s.size();
      } else {
        for (long long int i = 0; i < s.length() - 1; i++) {
          if ((s.at(i) == '0' && s.at(i + 1) == '1') ||
              (s.at(i) == '1' && s.at(i + 1) == '0')) {
            s.at(i) = '*';
            s.at(i + 1) = '*';
            mex += 2;
          }
        }
        mex += count(s.begin(), s.end(), '0');
      }
    }
    cout << mex << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  