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:     /*
       * To change this license header, choose License Headers in Project Properties.
       * To change this template file, choose Tools | Templates
       * and open the template in the editor.
       */
      import java.util.*;
      import java.util.Arrays;
  
  
      /**
       *
       * @author DELL
       */
        public class solution{
  
                 static long power(long n,long ans){
                     if(n==0)return ans;
                     return ans=power(n-1,ans)*2;
                 }
                 static boolean prime(int n){
                     for(int i=2;i*i<=n;i++){
                         if(n%i==0)return false;
                     }
                     return true;
                 }
                 static long power(long x, int y, int p)
                  {
                    long res = 1; // Initialize result
  
                    x = x % p; // Update x if it is more than or
                    // equal to p
  
                    if (x == 0)
                      return 0; // In case x is divisible by p;
  
                    while (y > 0)
                    {
  
                      // If y is odd, multiply x with result
                      if ((y & 1) != 0)
                        res = (res * x) % p;
  
                      // y must be even now
                      y = y >> 1; // y = y/2
                      x = (x * x) % p;
                    }
                    return res;
                  }
                 static int gcd(int a,int b)
                  {
                      if(b%a==0)
                      return a;
                      return gcd(b%a,a);
                  }
  
          public static void main(String[] args) {
               Scanner sc=new Scanner(System.in);
               StringBuilder sb=new StringBuilder();
               int t=sc.nextInt();
               while(t-->0){
                   int n=sc.nextInt();
                   String a=sc.next();
                   String b=sc.next();
                   int ans=0;
                   int arr[]=new int[n];
                   for(int i=0;i<a.length();i++){
                       if(a.charAt(i)!=b.charAt(i))ans+=2;
                       else if(a.charAt(i)=='1'){
                           if(i-1>=0 && arr[i-1]==0 && a.charAt(i-1)=='0' && b.charAt(i-1)=='0'){
                               ans++;
                               arr[i-1]=1;
                           }
                           else if(i+1<a.length() && a.charAt(i+1)=='0' && b.charAt(i+1)=='0' && arr[i+1]==0){
                               ans++;
                               arr[i+1]=1;
                           }
                       }
                       else ans++;
                   }
                   System.out.println(ans);
               }
          }
  
      }
  