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 java.io.*;
  import java.util.*;
  
  public class A {
      static final Reader s = new Reader();
      static final PrintWriter out = new PrintWriter(System.out);
  
      public static void main(String[] args) throws IOException {
          int t  = s.nextInt();
          for(int i=1; i<=t; ++i) {
  //            out.print("Case #"+i+": ");
              new Solver();
          }
          out.close();
      }
      static class Solver {
          Solver(){
              int n = s.nextInt();
              int[][] mat = new int[2][n];
              int b=0;
              long cnt=0;
              int last_index=-4;
              int value=4;
              for(int i=0;i<2;i++){
                  String temp=s.next();
                  char[] g = temp.toCharArray();
                  for(int j=0;j<n;j++){
                      mat[b][j] = (int)(g[j]-'0');
                  }
                  b++;
              }
  
  
              for(int i=0;i<n;i++){
                  if(mat[1][i]!=mat[0][i]){
                      cnt+=2;
                  }
                  else{
                      if(mat[0][i]==0){
                          if(last_index==i-1 && value==1){
                              cnt+=2;
                              value=9;
                          }
                          else {
                              cnt++;
                              value=0;
                          }
                          last_index=i;
                      }
                      else {
                          if(last_index==i-1){
                              if(value==0){
                                  cnt++;
                                  value=9;
                              }
                              else value=1;
                          }
                          else value=1;
                          last_index=i;
                      }
                  }
              }
              out.println(cnt);
          }
      }
      static class Reader {
          BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st;
          String next() {
              while(st==null||!st.hasMoreTokens()) {
                  try {
                      st=new StringTokenizer(in.readLine());
                  } catch(Exception e) {}
              }
              return st.nextToken();
          }
          int nextInt() {
              return Integer.parseInt(next());
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  
  }