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;
  using ll = long long;
  using Vi = vector<int>;
  using Pii = pair<int, int>;
  template <class t, class u, class... w>
  auto operator<<(t& o, u a) -> decltype(a.print(), o) {
    o << '{';
    a.print();
    return o << '}';
  }
  template <class t, class u, class... w>
  auto operator<<(t& o, u a) -> decltype(a.second, o) {
    o << '(' << a.first << ", " << a.second;
    return o << ')';
  }
  template <class t, class u, class... w>
  auto operator<<(t& o, u a) -> decltype((a).begin(), (a).end(), o) {
    o << '[';
    auto d = "";
    for (auto i : a) (o << d << i, d = ", ");
    return o << ']';
  }
  void DD(...) {}
  template <class t, class u, class... w>
  void DD(t s, u a, w... k) {
    for (int b = 1; *s && *s - b * 44; cerr << *s++) b += 2 / (*s * 2 - 81);
    cerr << ": " << a << *s++;
    DD(s, k...);
  }
  void run() {
    int n;
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    s1 = "#" + s1;
    s2 = "#" + s2;
    vector<int> dp(n + 1);
    int last[2];
    last[0] = -1;
    last[1] = -1;
    for (int i = 1; i <= n; i++) {
      last[s1[i] - '0'] = i;
      last[s2[i] - '0'] = i;
      if (last[0] == -1) {
        dp[i] = dp[i - 1];
      } else if (last[1] == -1) {
        dp[i] = dp[i - 1] + 1;
      } else if (last[0] < last[1]) {
        dp[i] = max(dp[last[0] - 1] + 2, dp[i - 1]);
      } else if (last[1] < last[0]) {
        dp[i] = max(dp[last[1] - 1] + 2, dp[i - 1] + 1);
      } else {
        dp[i] = dp[i - 1] + 2;
      }
    }
    cout << dp[n] << '\n';
  }
  int main() {
    cin.sync_with_stdio(0);
    cin.tie(0);
    cout << fixed << setprecision(10);
    int t;
    cin >> t;
    for (int i = 1; i <= t; i++) {
      run();
    }
    return 0;
  }
  