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: //package codeforce800;
  
  import java.util.Scanner;
  /*
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   * ��Ϊ0����00������1������������2���ϲ�
   * ��Ϊ2����11������0������������0���ϲ�
   * ���ϲ����У������бطָ�
   * */
  public class Max_Mex {
  	static int mex(int i,int j) {
  		if(i+j==2) return 0;
  		else return i+j+1;
  	}
  	
  	public static void main(String[] args) {
  		Scanner in=new Scanner(System.in);
  		int times=in.nextInt();
  		for (int i=1;i<=times;i++) {
  			int length=in.nextInt();
  			in.nextLine();
  			int sum=0;
  			char[] up=in.nextLine().toCharArray();
  			char[] down=in.nextLine().toCharArray();
  			for(int j=0;j<=length-1;) {
  				int mex=mex((int)up[j]-48,(int)down[j]-48);
  				if(j==length-1) {
  					sum+=mex;
  					break;
  				}
  				if(mex==2) {
  					sum+=2;
  					j++;
  				}else if(mex==1){
  					if((int)up[j+1]-48+(int)down[j+1]-48==2) {
  						sum+=2;
  						j+=2;
  					}else {
  						sum++;
  						j++;
  					}
  				}else {
  					if((int)up[j+1]-48+(int)down[j+1]-48==0) {
  						sum+=2;
  						j+=2;
  					}else j++;
  				}
  			}
  			System.out.println(sum);
  		}
  	}
  }
  