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;
  template <typename... T>
  void show(T&&... args) {
    ((cerr << args << " "), ...);
  }
  template <typename F, typename S>
  ostream& operator<<(ostream& os, const pair<F, S>& p) {
    return os << "(" << p.first << ", " << p.second << ")";
  }
  template <typename T>
  ostream& operator<<(ostream& os, const vector<T>& v) {
    os << "{";
    typename vector<T>::const_iterator it;
    for (it = v.begin(); it != v.end(); it++) {
      if (it != v.begin()) os << ", ";
      os << *it;
    }
    return os << "}";
  }
  template <typename T>
  ostream& operator<<(ostream& os, const set<T>& v) {
    os << "[";
    typename set<T>::const_iterator it;
    for (it = v.begin(); it != v.end(); it++) {
      if (it != v.begin()) os << ", ";
      os << *it;
    }
    return os << "]";
  }
  template <typename T>
  ostream& operator<<(ostream& os, const multiset<T>& v) {
    os << "[";
    typename multiset<T>::const_iterator it;
    for (it = v.begin(); it != v.end(); it++) {
      if (it != v.begin()) os << ", ";
      os << *it;
    }
    return os << "]";
  }
  template <typename F, typename S>
  ostream& operator<<(ostream& os, const map<F, S>& v) {
    os << "[";
    typename map<F, S>::const_iterator it;
    for (it = v.begin(); it != v.end(); it++) {
      if (it != v.begin()) os << ", ";
      os << it->first << " = " << it->second;
    }
    return os << "]";
  }
  template <typename F, typename S>
  ostream& operator<<(ostream& os, const unordered_map<F, S>& v) {
    os << "[";
    typename unordered_map<F, S>::const_iterator it;
    for (it = v.begin(); it != v.end(); it++) {
      if (it != v.begin()) os << ", ";
      os << it->first << " = " << it->second;
    }
    return os << "]";
  }
  template <typename... T>
  void read(T&... args) {
    ((cin >> args), ...);
  }
  template <typename... T>
  void write(T&&... args) {
    ((cout << args << " "), ...);
  }
  int solve() {
    int n;
    string str[2];
    cin >> n >> str[0] >> str[1];
    int ans = 0;
    for (int i = 0; i < n; i++) {
      int cnt = 0;
      if (str[0][i] == '0') cnt++;
      if (str[1][i] == '0') cnt++;
      if (cnt == 2) {
        if (i + 1 < n && str[0][i + 1] == '1' && str[1][i + 1] == '1') {
          i++;
          ans += 2;
        } else
          ans++;
      } else if (cnt == 0) {
        if (i + 1 < n && str[0][i + 1] == '0' && str[1][i + 1] == '0') {
          i++;
          ans += 2;
        }
      } else
        ans += 2;
    }
    cout << ans << "\n";
    return 0;
  }
  int main() {
    ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
    ;
    ;
    int t = 1, cs = 0;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  