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.BufferedReader;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  import java.util.*;
  import java.io.*;
  public class Main {
      public static void main(String args[])throws Exception{
          Input sc=new Input();
          StringBuilder sb=new StringBuilder();
          int t=sc.readInt();
          for(int f=0;f<t;f++){
              int n=sc.readInt();
              String str[]=new String[2];
              str[0]=sc.readString();
              str[1]=sc.readString();
              int arr[]=new int[n];
              for(int i=0;i<n;i++){
                  int v1=str[0].charAt(i)-'0';
                  int v2=str[1].charAt(i)-'0';
                  arr[i]=v1+v2;
              }
              int sum=0;
  //            for(int i=0;i<n;i++){
  //                System.out.print(arr[i]+" ");
  //            }
  //            System.out.println();
              for(int i=0;i<n;i++){
                  if(arr[i]==0){
                      if(i>=1 && arr[i-1]==2)
                          arr[i-1]=-1;
                      else if(i<n-1 && arr[i+1]==2)
                          arr[i+1]=-1;
                  }
              }
              for(int i=0;i<n;i++){
                  if(arr[i]==0)
                      arr[i]=1;
                  else if(arr[i]==1)
                      arr[i]=2;
                  else if(arr[i]==-1)
                      arr[i]=1;
                  else if(arr[i]==2)
                      arr[i]=0;
                  sum+=Math.abs(arr[i]);
              }
              if(arr[n-1]==-1)
                  sum++;
              sb.append(sum+"\n");
          }
          System.out.print(sb);
      }
  }
  class Input{
      BufferedReader br;
      StringTokenizer st;
      Input(){
          br=new BufferedReader(new InputStreamReader(System.in));
          st=new StringTokenizer("");
      }
      public int[] readArray() throws Exception{
          st=new StringTokenizer(br.readLine());
          int a[]=new int[st.countTokens()];
          for(int i=0;i<a.length;i++){
              a[i]=Integer.parseInt(st.nextToken());
          }
          return a;
      }
      public long[] readArrayLong() throws Exception{
          st=new StringTokenizer(br.readLine());
          long a[]=new long[st.countTokens()];
          for(int i=0;i<a.length;i++){
              a[i]=Long.parseLong(st.nextToken());
          }
          return a;
      }
      public int readInt() throws Exception{
          st=new StringTokenizer(br.readLine());
          return Integer.parseInt(st.nextToken());
      }
      public long readLong() throws Exception{
          st=new StringTokenizer(br.readLine());
          return Long.parseLong(st.nextToken());
      }
      public String readString() throws Exception{
          return br.readLine();
      }
      public int[][] read2dArray(int n,int m)throws Exception{
          int a[][]=new int[n][m];
          for(int i=0;i<n;i++){
              st=new StringTokenizer(br.readLine());
              for(int j=0;j<m;j++){
                  a[i][j]=Integer.parseInt(st.nextToken());
              }
          }
          return a;
      }
  }
  
  
  