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;
  void KruNULL() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  bool sortbysec(const pair<long long, long long> &a,
                 const pair<long long, long long> &b) {
    return (a.second < b.second);
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
  long long prime(long long n) {
    long long flag = 1;
    for (long long i = 2; i <= sqrt(n); i++) {
      if (n % i == 0) {
        flag = 0;
        break;
      }
    }
    return flag;
  }
  long long power(long long x, long long y, long long p) {
    long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  void solve() {
    long long n;
    cin >> n;
    string s, t;
    cin >> s >> t;
    long long dono = 0;
    map<char, long long> m;
    long long zz = 0;
    long long ans = 0;
    for (long long i = 0; i < n; i++) {
      m[s[i]]++;
      m[t[i]]++;
      if (s[i] == '0' && t[i] == '0') zz++;
      if (s[i] != t[i]) {
        ans += (2ll + zz);
        zz = 0;
        m.clear();
      } else if ((long long)m.size() == 2) {
        ans += max({2ll, zz, 2ll + zz - 1});
        zz = 0;
        m.clear();
      }
    }
    ans += zz;
    cout << ans << endl;
  }
  int32_t main() {
    KruNULL();
    long long tc = 1;
    cin >> tc;
    for (long long i = 1; i <= tc; i++) {
      solve();
    }
  }
  