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;
  inline long long read() {
    long long s = 0, w = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') w = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
    return s * w;
  }
  const long long p = 998244353;
  long long qp(long long x, long long y) {
    long long res = 1;
    for (long long t = x; y; y >>= 1, t = t * t % p)
      if (y & 1) res = res * t % p;
    return res;
  }
  char s[1000003], t[1000003];
  long long a[1000003];
  signed main() {
    for (long long T = read(); T--;) {
      long long n = read();
      scanf("%s", s + 1);
      scanf("%s", t + 1);
      for (long long i = 1; i <= n; ++i) a[i] = (s[i] == '1') + (t[i] == '1');
      long long cur = 0, qwq = 0, ans = 0;
      for (long long i = 1; i <= n; ++i)
        if (a[i] == 1) {
          ans += cur;
          ans += 2;
          cur = 0;
          qwq = 0;
        } else if (a[i] == 0) {
          ans += cur;
          if (qwq)
            qwq = 0, ans += 2, cur = 0;
          else
            cur = 1;
        } else {
          if (cur == 1)
            ans += 2, cur = 0;
          else
            qwq = 1;
        }
      printf("%lld\n", ans + cur);
    }
    return 0;
  }
  