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: import static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.*;
  
  public class A{
      public static void main(String[] args) throws IOException {
          FastScanner s = new FastScanner();
          int t = s.nextInt();
  //        int t = 1;
          while (t-- > 0) {
              int n = s.nextInt();
              int[][] a = new int[2][n];
              int k=0;
              for(int i=0;i<2;i++){
                  char[] g = s.next().toCharArray();
                  for(int j=0;j<n;j++){
                      a[k][j] = (int)(g[j]-'0');
                  }
                  k++;
              }
              long ans=0;
              int last=-4;
              int v=4;
              for(int i=0;i<n;i++){
                  if(a[0][i]!=a[1][i]){
                      ans+=2;
                  }
                  else{
                      if(a[0][i]==0){
                          if(last==i-1 && v==1){
                              ans+=2;
                              v=4;
                          }
                          else {
                              ans++;
                              v=0;
                          }
                          last=i;
                      }
                      else {
                          if(last==i-1){
                              if(v==0){
                                  ans++;
                                  v=4;
                              }
                              else v=1;
                          }
                          else v=1;
                          last=i;
                      }
                  }
              }
              out.println(ans);
          }
      }
      static long gcd(long a, long b) throws IOException {
          return (b == 0) ? a : gcd(b, a % b);
      }
  
      static int gcd(int a, int b) throws IOException {
          return (b == 0) ? a : gcd(b, a % b);
      }
  
      static void sortr(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i : a) l.add(i);
          Collections.sort(l, Collections.reverseOrder());
          for (int i = 0; i < a.length; i++) a[i] = l.get(i);
      }
  
      static void sort(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i : a) l.add(i);
          Collections.sort(l);
          for (int i = 0; i < a.length; i++) a[i] = l.get(i);
      }
  
      static void sort(long[] a) {
          ArrayList<Long> l = new ArrayList<>();
          for (long i : a) l.add(i);
          Collections.sort(l);
          for (int i = 0; i < a.length; i++) a[i] = l.get(i);
      }
  
      static class FastScanner {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
  
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] readArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  }
  