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;
  const int iinf = 1e9 + 7;
  const long long int inf = 1e9 + 7;
  const long long int mod = 998244353;
  const long long int maxn = 2e5 + 5;
  long long int sqr(long long int x) { return x * x; }
  long long int modd(long long int a) {
    if (a > 0) return a % inf;
    while (a < 0) {
      a += inf;
    }
    return a;
  }
  long long int modexpo(long long int a, long long int b, long long int iinf1) {
    a %= iinf1;
    long long int res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % iinf1;
      a = a * a % iinf1;
      b >>= 1;
    }
    return res;
  }
  void solve() {
    int n;
    cin >> n;
    string s, s1;
    cin >> s >> s1;
    long long int ans = 0;
    vector<bool> vis(n, false);
    for (int i = 0; i < n - 1; i++) {
      if (s[i] != s1[i]) {
        ans += 2;
        vis[i] = true;
      } else {
        if (s[i + 1] == s1[i + 1] && s[i] != s[i + 1]) {
          vis[i] = true;
          vis[i + 1] = true;
          ans += 2;
          i++;
        } else {
          if (s[i] == '0') {
            ans += 1;
          }
          vis[i] = true;
        }
      }
    }
    if (!vis[n - 1]) {
      if (s[n - 1] != s1[n - 1]) {
        ans += 2;
      } else {
        if (s[n - 1] == '0') ans++;
      }
    }
    cout << ans << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  