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 INF = ((1 << 30) - 1);
  int dirs[4][2] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
  void split(string &line, vector<string> &v, const char delimeter = ',') {
    stringstream str(line);
    string buffer;
    while (getline(str, buffer, delimeter)) {
      v.push_back(buffer);
    }
  }
  void test(string ans_file, string guess_file) {
    ifstream ans(ans_file);
    ifstream guess(guess_file);
    string a, g;
    int line = 1;
    while (!ans.eof()) {
      if (guess.eof()) {
        printf("There is no guess. file[%s] line : [%d]\n", ans_file.c_str(),
               line);
        exit(-1);
      }
      ans >> a;
      guess >> g;
      if (a != g) {
        printf("wrong guess! file[%s] line : [%d]\n", ans_file.c_str(), line);
        exit(-1);
      }
      line++;
    }
    if (!guess.eof()) {
      printf("There is extra guess. file[%s] line : [%d]\n", ans_file.c_str(),
             line);
      exit(-1);
    }
  }
  int n, s;
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0), cout.precision(15);
    int t;
    cin >> t;
    while (t-- > 0) {
      int n;
      cin >> n;
      string s1, s2;
      cin >> s1 >> s2;
      int sum = 0;
      int prev = -1;
      for (int i = 0; i < n; i++) {
        int a = s1[i] - '0', b = s2[i] - '0';
        int set = (1 << a) | (1 << b);
        if (set == 3) {
          sum += 2;
          if (prev == 0) {
            sum += 1;
          }
          prev = -1;
        } else if (set == 1) {
          if (prev == 1) {
            sum += 2;
            prev = -1;
          } else if (prev == 0) {
            sum += 1;
            prev = 0;
          } else {
            prev = 0;
          }
        } else {
          if (prev == -1)
            prev = 1;
          else if (prev == 0) {
            sum += 2;
            prev = -1;
          }
        }
      }
      if (prev == 0) sum++;
      cout << sum << '\n';
    }
  }
  