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 precision(int a) { cout << setprecision(a) << fixed; }
  vector<int> sieve(int n) {
    int* arr = new int[n + 1]();
    vector<int> vect;
    for (int i = 2; i <= n; i++)
      if (arr[i] == 0) {
        vect.push_back(i);
        for (long long j = (long long(i) * long long(i)); j <= n; j += i)
          arr[j] = 1;
      }
    return vect;
  }
  static bool comp(const pair<long long, long long>& a,
                   const pair<long long, long long>& b) {
    return a.second < b.second;
  }
  const long long N = 3e5 + 7;
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      vector<string> v;
      string s1, s2;
      cin >> s1 >> s2;
      v.push_back(s1);
      v.push_back(s2);
      long long cnt0 = 0, cnt1 = 0;
      for (long long i = 0; i < (2); i++) {
        for (long long j = 0; j < (n); j++) {
          if (v[i][j] == '0')
            cnt0++;
          else
            cnt1++;
        }
      }
      if (cnt0 == 0)
        cout << 0 << "\n";
      else if (cnt1 == 0)
        cout << n << "\n";
      else {
        long long cnt2 = 0, cnt3 = 0, cnt4 = 0, cnt5 = 0;
        for (long long i = 0; i < (n); i++) {
          if ((v[0][i] == '0' && v[1][i] == '1') ||
              (v[0][i] == '1' && v[1][i] == '0'))
            cnt2++;
          else if (v[0][i] == '0' && v[1][i] == '0') {
            if (i < n - 1 && v[0][i + 1] == '1' && v[1][i + 1] == '1') {
              cnt3++;
              i++;
            } else
              cnt4++;
          } else if (v[0][i] == '1' && v[1][i] == '1') {
            if (i < n - 1 && v[0][i + 1] == '0' && v[1][i + 1] == '0') {
              cnt5++;
              i++;
            }
          }
        }
        cout << 2 * cnt2 + 2 * cnt3 + cnt4 + 2 * cnt5 << "\n";
      }
    }
    return 0;
  }
  