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>
  long long q_pow(long long x, long long n, long long mod) {
    long long ret = 1;
    while (n) {
      if (n & 1) ret = x * ret % mod;
      x = x * x % mod;
      n >>= 1;
    }
    return ret;
  }
  long long getbit(long long now) {
    int cnt = 0;
    while (now) {
      cnt++;
      now -= (now) & (-now);
    }
    return cnt;
  }
  int dx[] = {-1, 1, 0, 0};
  int dy[] = {0, 0, -1, 1};
  using namespace std;
  const long long maxn = 1e5 + 10;
  const long long INF = 0x3f3f3f3f;
  const long long IINF = 0x3f3f3f3f3f3f3f3f;
  const double eps = 1e-9;
  const long long mod = 998244353;
  void add(long long &x, long long y) {
    x += y;
    if (x >= mod) x -= mod;
  }
  void mul(long long &x, long long y) {
    x -= y;
    if (x < 0) x += mod;
  }
  char s1[maxn];
  char s2[maxn];
  void solve() {
    int n;
    cin >> n;
    cin >> (s1 + 1);
    cin >> (s2 + 1);
    int ans = 0;
    int cnt0 = 0, cnt1 = 0;
    for (int i = 1; i <= n; i++) {
      s1[i] -= '0';
      s2[i] -= '0';
    }
    for (int i = 1; i <= n; i++) {
      if (s1[i] == 0 or s2[i] == 0) {
        if (cnt0) ans++;
        cnt0 = 1;
      }
      if (s1[i] == 1 or s2[i] == 1) {
        cnt1 = 1;
      }
      if (cnt1 and cnt0) ans += 2, cnt1 = cnt0 = 0;
    }
    if (cnt0) ans++;
    cout << ans << "\n";
  }
  int main() {
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  