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;
  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 - 48;
      ch = getchar();
    }
    return x * f;
  }
  int t;
  char r1[1000001];
  char r2[1000001];
  int typ[1000001];
  int fg[1000001];
  int main() {
    cin >> t;
    while (t--) {
      int n;
      n = read();
      cin >> r1 + 1;
      cin >> r2 + 1;
      for (int i = 0; i <= n + 6; i++) {
        fg[i] = 0;
        typ[i] = 0;
      }
      for (int i = 1; i <= n; i++) {
        if (r1[i] == '0' && r2[i] == '0') {
          typ[i] = 1;
        } else {
          if (r1[i] == '1' && r2[i] == '1') {
            typ[i] = 3;
          } else {
            typ[i] = 2;
          }
        }
      }
      int ans = 0;
      for (int i = 1; i <= n; i++) {
        if (typ[i] == 1) {
          if (typ[i - 1] == 3 && fg[i - 1] == 0) {
            ans += 2;
            fg[i - 1] = 1;
          } else {
            if (typ[i + 1] == 3 && fg[i + 1] == 0) {
              ans += 2;
              fg[i + 1] = 1;
            } else {
              ans++;
            }
          }
        } else {
          if (typ[i] == 2) {
            ans += 2;
          } else {
            ans += 0;
          }
        }
      }
      cout << ans << "\n";
    }
    return 0;
  }
  