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;
  int gcd(int a, int b);
  bool prime(int n);
  long long sumdigit(long long n);
  bool ispowerof2(int n);
  long long onesinbinary(long long n);
  long long power(long long a, long long b);
  bool cmp(pair<int, int> x, pair<int, int> y) { return x.second < y.second; }
  bool cmp(int x, int y) { return x > y; }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      char arr[2][n];
      for (int i = 0; i < 2; i++) {
        for (int j = 0; j < n; j++) {
          cin >> arr[i][j];
        }
      }
      int ans = 0;
      vector<int> v;
      for (int i = 0; i < n; i++) {
        if (arr[0][i] == '0' && arr[1][i] == '0')
          ans++;
        else if (arr[0][i] == '0' || arr[1][i] == '0')
          ans += 2;
        else {
          v.push_back(i);
        }
      }
      int x = v.size();
      for (int i = 0; i < x; i++) {
        if (v[i] < n - 1 && v[i] > 0) {
          if (arr[0][v[i] - 1] == '0' && arr[1][v[i] - 1] == '0') {
            ans++;
          } else if (arr[0][v[i] + 1] == '0' && arr[1][v[i] + 1] == '0') {
            ans++;
            arr[0][v[i] + 1] = '1', arr[1][v[i] + 1] = '1';
          }
        } else if (v[i] == n - 1) {
          if (arr[0][v[i] - 1] == '0' && arr[1][v[i] - 1] == '0') {
            ans++;
          }
        } else if (v[i] == 0) {
          if (arr[0][v[i] + 1] == '0' && arr[1][v[i] + 1] == '0') {
            ans++;
            arr[0][v[i] + 1] = '1', arr[1][v[i] + 1] = '1';
          }
        }
      }
      cout << ans << endl;
    }
    return 0;
  }
  long long onesinbinary(long long n) {
    long long cnt = 0;
    while (n) {
      n = n & (n - 1);
      cnt++;
    }
    return cnt;
  }
  int gcd(int a, int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  bool prime(int n) {
    for (int i = 2; i * i <= n; i++)
      if (n % i == 0) return false;
    return true;
  }
  long long sumdigit(long long n) {
    int c = 0;
    while (n > 0) {
      c += n % 10;
      n /= 10;
    }
    return c;
  }
  bool ispowerof2(int n) {
    if (n == 0) return false;
    return (!(n & ((n) & (n - 1))));
  }
  long long power(long long a, long long b) {
    if (b == 0) return 1;
    long long res = 1;
    while (b) {
      if (b % 2 == 1) res = (res * a) % 1000000007;
      a = (a * a) % 1000000007;
      b /= 2;
    }
    return res;
  }
  