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;
  #pragma comment(linker, "/STACK:102400000,102400000")
  long long mul(long long a, long long b) { return (a * b) % (1000000007); }
  long long add(long long a, long long b) { return (a + b) % (1000000007); }
  long long sub(long long a, long long b) {
    return ((a - b) % (1000000007) + (1000000007)) % (1000000007);
  }
  void upd(long long &a, long long b) {
    a = (a % (1000000007) + b % (1000000007)) % (1000000007);
  }
  inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  string s, s2;
  int n;
  int mex(int l, int r) {
    bool fl0 = 0, fl1 = 0;
    for (int i = l; i <= r; i++) {
      if (s[i] == '0')
        fl0 = 1;
      else
        fl1 = 1;
      if (s2[i] == '0')
        fl0 = 1;
      else
        fl1 = 1;
    }
    if (!fl0)
      return 0;
    else if (!fl1)
      return 1;
    else
      return 2;
  }
  int f[512345];
  int main() {
    int T = read();
    while (T--) {
      n = read();
      cin >> s >> s2;
      int n = s.length();
      f[0] = mex(0, 0);
      for (int i = 1; i <= n - 1; i++) {
        f[i] = f[i - 1] + mex(i, i);
        if (i > 1) f[i] = max(f[i], f[i - 2] + mex(i - 1, i));
        if (i == 1) f[i] = max(f[i], mex(i - 1, i));
      }
      cout << f[n - 1] << endl;
    }
    return 0;
  }
  