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.util.*;
  import java.io.*;
  
  public class C_MAX_MEX_Cut{
      public static void main(String[] args) {
          FastScanner s= new FastScanner();
          StringBuilder res = new StringBuilder();
          int t=s.nextInt();
          int p=0;
          while(p<t){
          int n=s.nextInt();
          String str1=s.nextToken();
          String str2=s.nextToken();
          int array[]= new int[n];
          for(int i=0;i<n;i++){
              char ch1=str1.charAt(i);
              char ch2=str2.charAt(i);
              int num1=ch1-'0';
              int num2=ch2-'0';
              array[i]=num1+num2;
          }
          long ans=0;
          for(int i=0;i<n;i++){
              if(array[i]==1){
                  ans+=2;
              }
              else if(array[i]==0){
                  if((i+1)<n){
                      if(array[i+1]==2){
                          ans+=2;
                          i++;
                      }
                      else{
                          ans+=1;
                      }
                  }
                  else{
                      ans+=1;
                  }
              }
              else{
                  if((i+1)<n){
                      if(array[i+1]==0){
                          ans+=2;
                          i++;
                      }
                  }
  
              }
          }
          res.append(ans+"\n");
          p++;
      }
      System.out.println(res);
      }
      static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
   
  		public FastScanner(String s) {
  			try {
  				br = new BufferedReader(new FileReader(s));
  			} catch (FileNotFoundException e) {
  				// TODO Auto-generated catch block
  				e.printStackTrace();
  			}
  		}
   
  		public FastScanner() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
   
  		String nextToken() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					// TODO Auto-generated catch block
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
   
  		int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
   
  		long nextLong() {
  			return Long.parseLong(nextToken());
  		}
   
  		double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  	}
  
  
  
  
  }