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 maxn = 2e5 + 10;
  const long double scale = 1e+9;
  long long gcd(long long a, long long b) {
    if (b > a) {
      return gcd(b, a);
    }
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  long long expo(long long a, long long b, long long mod) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b = b >> 1;
    }
    return res;
  }
  long long mminvprime(long long a, long long b) { return expo(a, b - 2, b); }
  bool revsort(long long a, long long b) { return a > b; }
  void swap(int &x, int &y) {
    int temp = x;
    x = y;
    y = temp;
  }
  long long combination(long long n, long long r, long long m, long long *fact,
                        long long *ifact) {
    long long val1 = fact[n];
    long long val2 = ifact[n - r];
    long long val3 = ifact[r];
    return (((val1 * val2) % m) * val3) % m;
  }
  void google(int t) { cout << "Case #" << t << ": "; }
  vector<long long> sieve(int n) {
    int *arr = new int[n + 1]();
    vector<long long> vect;
    for (int i = 2; i <= n; i++)
      if (arr[i] == 0) {
        vect.push_back(i);
        for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
      }
    return vect;
  }
  long long mod_add(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long mod_mul(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long mod_sub(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  long long mod_div(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (mod_mul(a, mminvprime(b, m), m) + m) % m;
  }
  long long phin(long long n) {
    long long number = n;
    if (n % 2 == 0) {
      number /= 2;
      while (n % 2 == 0) n /= 2;
    }
    for (long long i = 3; i <= sqrt(n); i += 2) {
      if (n % i == 0) {
        while (n % i == 0) n /= i;
        number = (number / i * (i - 1));
      }
    }
    if (n > 1) number = (number / n * (n - 1));
    return number;
  }
  bool isPrime(long long n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (long long i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  long long nextPrime(long long N) {
    if (N <= 1) return 2;
    long long prime = N;
    bool found = false;
    while (!found) {
      prime++;
      if (isPrime(prime)) found = true;
    }
    return prime;
  }
  bool palin(string s, long long i, long long j) {
    if (i > j) return false;
    while (i < j) {
      if (s[i] != s[j]) return false;
      i++;
      j--;
    }
    return true;
  }
  bool isperfect(long double x) {
    if (x >= 0) {
      long long s = sqrt(x);
      return (s * s == x);
    }
    return false;
  }
  bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
    return a.first < b.first;
  }
  long long power(long long num, long long p, long long m) {
    long long x = 1;
    while (p > 0) {
      if (p & 1) {
        x = (x * num) % m;
        num--;
      }
      num = (num * num) % m;
      p /= 2;
    }
    return x % m;
  }
  vector<long long> factor(long long x) {
    vector<long long> res;
    for (long long i = 2; i * i <= x; i++) {
      if (x % i == 0) res.push_back(i);
      while (x % i == 0) {
        x /= i;
      }
    }
    if (x != 1) res.push_back(x);
    return res;
  }
  long long inverseFact(long long n) {
    return power(n, 998244353 - 2, 998244353);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      string s, t;
      cin >> s >> t;
      long long i = 0;
      long long res = 0, res2 = 0;
      while (i < n) {
        if ((s[i] == '1' && t[i] == '0') || (t[i] == '1' && s[i] == '0')) {
          res += 2;
          i++;
          continue;
        } else if ((s[i] == '1' && t[i] == '1')) {
          if (i == n - 1) break;
          if ((s[i + 1] == '0' && t[i + 1] == '1') ||
              (s[i + 1] == '1' && t[i + 1] == '0')) {
            i += 2;
            res += 2;
          } else if (s[i + 1] == '0' && t[i + 1] == '0') {
            i += 2;
            res += 2;
          } else
            i++;
        } else {
          if (i == n - 1) {
            res += 1;
            break;
          }
          if (s[i + 1] == '1' && t[i + 1] == '1') {
            res += 2;
            i += 2;
          } else {
            res += 1;
            i++;
          }
        }
      }
      reverse(s.begin(), s.end());
      reverse(t.begin(), t.end());
      i = 0;
      while (i < n) {
        if ((s[i] == '1' && t[i] == '0') || (t[i] == '1' && s[i] == '0')) {
          res2 += 2;
          i++;
        } else if ((s[i] == '1' && t[i] == '1')) {
          if (i == n - 1) break;
          if ((s[i + 1] == '0' && t[i + 1] == '1') ||
              (s[i + 1] == '1' && t[i + 1] == '0')) {
            i += 2;
            res2 += 2;
          } else if (s[i + 1] == '0' && t[i + 1] == '0') {
            i += 2;
            res2 += 2;
          } else
            i++;
        } else {
          if (i == n - 1) {
            res2 += 1;
            break;
          }
          if (s[i + 1] == '1' && t[i + 1] == '1') {
            res2 += 2;
            i += 2;
          } else {
            res2 += 1;
            i++;
          }
        }
      }
      cout << max(res2, res) << "\n";
    }
    return 0;
  }
  