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 org.omg.CORBA.IMP_LIMIT;
  
  import java.io.*;
  import java.math.BigInteger;
  import java.net.InetSocketAddress;
  import java.util.*;
  import java.math.BigDecimal;
  
  public class Division {
      public static void main(String[] args) throws IOException {
          PrintWriter pw = new PrintWriter(System.out);
         int t = nextInt();
          for (int i = 0; i < t; i++) {
              int a = nextInt();
              String str = nextToken();
              String s = nextToken();
              int sum = 0;
              int nol = 0;
              int ed = 0;
              for (int j = 0; j < str.length(); j++) {
                  if(str.charAt(j)-'0'==0){
                      nol++;
                  }else{
                      ed++;
                  }
                  if(s.charAt(j)-'0'==0){
                      nol++;
                  }else{
                      ed++;
                  }
                  if(nol==3 && ed==1){
                      sum+=3;
                      nol=0;
                      ed=0;
                      continue;
                  }
                  if(nol>=1 && ed>=1){
                      sum+=2;
                      nol=0;
                      ed=0;
                      continue;
                  }
                  if(ed==0) {
                      if (nol == 4) {
                          sum += 1;
                          nol = 2;
                          ed = 0;
                      } else {
                           continue;
                      }
                      continue;
                  }
              }
              if(nol==2 ){
                  sum++;
              }
              nol=0;
              ed=0;
              System.out.println(sum);
          }
      }
  
      static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      static PrintWriter out = new PrintWriter(System.out);
      static StringTokenizer in = new StringTokenizer("");
  
      public static boolean hasNext() throws IOException {
          if (in.hasMoreTokens()) return true;
          String s;
          while ((s = br.readLine()) != null) {
              in = new StringTokenizer(s);
              if (in.hasMoreTokens()) return true;
          }
          return false;
      }
  
      public static String nextToken() throws IOException {
          while (!in.hasMoreTokens()) {
              in = new StringTokenizer(br.readLine());
          }
          return in.nextToken();
      }
  
      public static int nextInt() throws IOException {
          return Integer.parseInt(nextToken());
      }
  
      public static double nextDouble() throws IOException {
          return Double.parseDouble(nextToken());
      }
  
      public static long nextLong() throws IOException {
          return Long.parseLong(nextToken());
      }
  }