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;
  static long long int mod = (long long)(1e9) + 7;
  long long power(long long x, long long y, long long p) {
    long long res = 1;
    x = x % p;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  vector<int> prime((long long)(1e7), 1);
  void seive(long long n) {
    long long i, j;
    prime[0] = prime[1] = 0;
    for (i = 2; i * i <= n; i++) {
      if (prime[i] == 1) {
        for (j = i * i; j <= n; j += i) prime[j] = 0;
      }
    }
  }
  void solve() {
    char ch[] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm',
                 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
    int n;
    cin >> n;
    string str1, str2;
    cin >> str1;
    cin >> str2;
    int i, ans = 0, zeros = 0;
    map<char, int> m1;
    for (i = 0; i < n - 1; i++) {
      map<char, int> m1;
      m1[str1[i]]++;
      m1[str2[i]]++;
      if (m1['0'] > 0 && m1['1'] > 0)
        ans += 2;
      else {
        m1.clear();
        m1[str1[i + 1]]++;
        m1[str2[i + 1]]++;
        if (str1[i] == '0' && str2[i] == '0' && m1['1'] == 2) {
          ans += 2;
          i++;
        } else if (str1[i] == '1' && str2[i] == '1' &&
                   (str1[i + 1] == '0' || str2[i + 1] == '0')) {
          ans += 2;
          i++;
        } else if (str1[i] == '0' && str2[i] == '0')
          ans++;
      }
    }
    if (i == n)
      ;
    else {
      map<char, int> m1;
      m1[str1[i]]++;
      m1[str2[i]]++;
      if (m1['0'] > 0 && m1['1'] > 0)
        ans += 2;
      else if (m1['0'] > 0)
        ans++;
    }
    cout << ans << "\n";
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  