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>
  #pragma GCC optimize("O3", "unroll-loops")
  #pragma GCC target("sse4.2")
  using namespace std;
  mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
  void flush() { cout << flush; }
  void flushln() { cout << endl; }
  void println() { cout << '\n'; }
  template <class T>
  void print(const T &x) {
    cout << x;
  }
  template <class T>
  void read(T &x) {
    cin >> x;
  }
  template <class T, class... U>
  void read(T &x, U &...u) {
    read(x);
    read(u...);
  }
  template <class T, class... U>
  void print(const T &x, const U &...u) {
    print(x);
    print(u...);
  }
  template <class T, class... U>
  void println(const T &x, const U &...u) {
    print(x);
    println(u...);
  }
  template <class T, class U>
  inline bool chmin(T &x, const U &y) {
    return y < x ? x = y, 1 : 0;
  }
  template <class T, class U>
  inline bool chmax(T &x, const U &y) {
    return y > x ? x = y, 1 : 0;
  }
  template <class T>
  inline void sort(T &a) {
    sort((a).begin(), (a).end());
  }
  template <class T>
  inline void rsort(T &a) {
    sort((a).rbegin(), (a).rend());
  }
  template <class T>
  inline void reverse(T &a) {
    reverse((a).begin(), (a).end());
  }
  template <class T, class U>
  inline istream &operator>>(istream &str, pair<T, U> &p) {
    return str >> p.first >> p.second;
  }
  template <class T>
  inline istream &operator>>(istream &str, vector<T> &a) {
    for (auto &i : a) str >> i;
    return str;
  }
  template <class T>
  inline T sorted(T a) {
    sort(a);
    return a;
  }
  void solve() {
    int n;
    read(n);
    string second, t;
    read(second, t);
    vector<int> dp(1 << 3, -1000000007);
    dp[0] = 0;
    auto getmex = [](int x) { return __builtin_ctz(x + 1); };
    for (int i = 0; i < n; ++i) {
      vector<int> newdp(1 << 3, -1000000007);
      int vs = second[i] - '0';
      int vt = t[i] - '0';
      for (int was = 0; was < (1 << 3); ++was) {
        chmax(newdp[was | (1 << vs) | (1 << vt)], dp[was]);
        chmax(newdp[(1 << vs) | (1 << vt)], dp[was] + getmex(was));
      }
      dp = move(newdp);
    }
    int ans = -1000000007;
    for (int i = 0; i < (1 << 3); ++i) chmax(ans, dp[i] + getmex(i));
    println(ans);
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(12);
    int t;
    read(t);
    while (t--) solve();
    return 0;
  }
  