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.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  	public class practice{
  		public static void main(String[] args){
  			FastScanner r = new FastScanner();					
  			int t = r.i() ,n;
  			char[][] mat;
  			int max=0;
  			boolean[] vis;
  			StringBuffer ans = new StringBuffer("");
  			while(t-->0) {
  				n = r.i();
  				mat  = new char[2][n];
  				mat[0] = r.n().toCharArray();
  				mat[1] = r.n().toCharArray();
  				max=0;
  				vis = new boolean[n];
  				for(int i=0;i<n;i++) {
  					if(mat[0][i] != mat[1][i]) {
  						max+=2;
  					}
  					else if(mat[0][i]=='0') {
  						
  						
  						if( i-1>=0&&!vis[i-1] && mat[0][i-1] == mat[1][i-1] && mat[1][i-1]=='1') {
  							vis[i]=true;
  						max+=2;}
  						else
  								max++;
  					}
  					
  					else {
  						if( i-1>=0&&!vis[i-1] && mat[0][i-1] == mat[1][i-1] && mat[1][i-1]=='0') {
  							max++;
  							vis[i] = true;
  							}
  					}
  //					p(max + " c");
  					
  				}
  				ans.append(max + "\n");
  			}				
  			p(ans);
  		}
  
  		
  		static <T> void p(T s) {
  			System.out.println(s);
  		}
  		
  		static void p() {
  			System.out.println();
  		}		
  		static <T> void pn(T s) {
  			System.out.print(s);
  		}
  
  	}
  		
  
  	
  	class FastScanner {
  	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  	    StringTokenizer st=new StringTokenizer("");
  	    String n() {
  	        while (!st.hasMoreTokens())
  	            try {
  	                st=new StringTokenizer(br.readLine());
  	            } catch (IOException e) {
  	                e.printStackTrace();
  	            }
  	        return st.nextToken();
  	    }
  	    int[] ia(int n) {
  	    	int[] arr = new int[n];
  	    	for(int i=0;i<n;i++)
  	    		arr[i] = i();
  	    	return arr;
  	    }
  	    String[] sa(int n) {
  	    	String[] arr = new String[n];
  	    	for(int i=0;i<n;i++)
  	    		arr[i] = n();
  	    	return arr;
  	    }
  	    long[] la(int n) {
  	    	long[] arr = new long[n];
  	    	for(int i=0;i<n;i++)
  	    		arr[i] = l();
  	    	return arr;
  	    }
  	    String nl()
  	    {
  	        String str = "";
  	        try
  	        {
  	            str = br.readLine();
  	        }
  	        catch (IOException e)
  	        {
  	            e.printStackTrace();
  	        }
  	        return str;
  	    }
  
  	    int i() {
  	        return Integer.parseInt(n());
  	    }
  	    long l() {
  	        return Long.parseLong(n());
  	    }
  	}
  