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 long long mod = 1e9 + 7;
  const long long inf = 1e18 + 9;
  const long long MAX_SIZE = 1e7 + 1;
  long long dx[4] = {0, -1, 1, 0}, dy[4] = {1, 0, 0, -1};
  long long _t = 1, _T;
  long long max(long long a, long long b) {
    if (a < b) return b;
    return a;
  }
  long long min(long long a, long long b) {
    if (a > b) return b;
    return a;
  }
  long long mmi(long long a, long long m) {
    long long m0 = m;
    long long y = 0, x = 1;
    if (m == 1) return 0;
    while (a > 1) {
      long long q = a / m;
      long long t = m;
      m = a % m, a = t;
      t = y;
      y = x - q * y;
      x = t;
    }
    if (x < 0) x += m0;
    return x;
  }
  long long nxor(long long n) {
    if (n % 4 == 0) return n;
    if (n % 4 == 1) return 1;
    if (n % 4 == 2) return n + 1;
    return 0;
  }
  long long sum(long long n) { return ((n * (n + 1)) / 2); }
  long long modsum(long long n, long long m) {
    long long ans = n % m;
    ans *= (n + 1) % m;
    ans %= m;
    ans *= mmi(2, m);
    return ans % m;
  }
  long long leap(long long y) {
    if (y % 400 == 0)
      return 1;
    else if (y % 100 == 0)
      return 0;
    else if (y % 4 == 0)
      return 1;
    else
      return 0;
  }
  long long mypow(long long x, long long n) {
    long long nn = n, ans = 1;
    while (nn) {
      if (nn % 2)
        ans *= x, nn--;
      else
        x *= x, nn /= 2;
    }
    return ans;
  }
  long long modpow(long long x, long long n, long long md) {
    long long nn = n, ans = 1;
    while (nn) {
      if (nn % 2)
        ans *= x, ans %= md, nn--;
      else
        x *= x, x %= md, nn /= 2;
    }
    return ans;
  }
  void solve(long long csn);
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(0);
    bool multicases = 1;
    if (multicases) cin >> _t;
    _T = _t;
    while (_t--) solve(_T - _t);
    return 0;
  }
  void solve(long long csn) {
    long long n;
    cin >> n;
    ;
    string s1, s2;
    cin >> s1 >> s2;
    long long ans = 0, i = 0;
    while (i < n) {
      if (s1[i] != s2[i]) {
        ans += 2;
        i++;
      } else if (s1[i] == '1') {
        if (i == n - 1)
          break;
        else if (s1[i + 1] == '0' || s2[i + 1] == '0') {
          ans += 2;
          i += 2;
        } else
          i++;
      } else {
        if (i == n - 1) {
          ans++;
          break;
        } else if (s1[i + 1] == '1' && s2[i + 1] == '1') {
          ans += 2;
          i += 2;
        } else {
          ans++;
          i++;
        }
      }
    }
    cout << ans << "\n";
    return;
  }
  