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.lang.*;
   import java.io.*;
  
   public class Main
   {
       static long M = 998244353;
       static final PrintWriter out =new PrintWriter(System.out);
       static final FastReader fs = new FastReader();
       static boolean prime[];
      
       public static void main (String[] args) throws java.lang.Exception
       {
             int t = fs.nextInt();
             while(t-->0)solve();
             out.flush();
      }
  
      static void solve() throws java.lang.Exception{
          // out.println("GG");
          int n = fs.nextInt();
          String s1= fs.next();
          String s2= fs.next();
  
          int count = 0;
          // int i = 0;
          int z =0;
          int o =0;
          int len = 0;
          for(int i=0;i<n;i++){
              if(s1.charAt(i)=='0'||s2.charAt(i)=='0')z++;
              if(s1.charAt(i)=='1'||s2.charAt(i)=='1')o++;
              if(z!=0&&o!=0){
                  if(z>o){
                      z = z-o;
                      count+=z;
                  }
                  count += 2;
                  z=0;
                  o=0;
              }
              // out.println(o+" "+z+"****");
          }
          int cz=0;
          for(int i=0;i<n;i++){
              int j = i;
              boolean f = false;
              int dd = 0;
              while(j<n && s1.charAt(j)=='0'&&s2.charAt(j)=='0'){
                  j++;
                  dd++;
                  f = true;
              }
  
              if(f){
                  i = j-1;
                  cz+=dd;
              }
          }
  
          if(z!=0){
              count+=z;
          }
  
          // out.println(cz+"**");
          out.println(Math.max(cz,count));
          
      }
    
    static long mod(long a)
    {
        return (a%M + M) % M;
    }
    static void sieveOfEratosthenes(int n)
      {
           prime = new boolean[n + 1];
          for (int i = 0; i <= n; i++)
              prime[i] = true;
              prime[0]=false;
              if(1<=n)
              prime[1]=false;
          for (int p = 2; p * p <= n; p++)
          {
              
              if (prime[p] == true)
              {
                 
                  for (int i = p * p; i <= n; i += p)
                      prime[i] = false;
              }
          }
      }
           static class FastReader 
           {
           BufferedReader br;
           StringTokenizer st;
    
           public FastReader()
           {
               br = new BufferedReader(
                   new InputStreamReader(System.in));
           }
    
           String next()
           {
               while (st == null || !st.hasMoreElements()) {
                   try {
                       st = new StringTokenizer(br.readLine());
                       }
                   catch (IOException e) {
                       e.printStackTrace();
                       }
           }
               return st.nextToken();
           }
    
           int nextInt() { return Integer.parseInt(next()); }
    
           long nextLong() { return Long.parseLong(next()); }
    
           double nextDouble()
           {
               return Double.parseDouble(next());
           }
    
           String nextLine()
           {
               String str = "";
               try {
                   str = br.readLine();
               }
               catch (IOException e) {
                   e.printStackTrace();
               }
               return str;
           }
   long [] arrayIn(int n) throws IOException
       {
           long  arr[] = new long[n];
           for(int i=0; i<n; i++)
           {
               arr[i] = nextLong();
           }
           return arr;
       }
       }
       
       public static class Pairs implements Comparable<Pairs>
       {
           int value,index;
           Pairs(int value, int index)
           {
               this.value = value;
               this.index = index;
           }
           public int compareTo(Pairs p)
           {
               return Integer.compare(this.value, p.value);
           }
       }
       static final Random random = new Random();
    
       static void ruffleSort(int arr[])
       {
           int n = arr.length;
           for(int i=0; i<n; i++)
           {
               int j = random.nextInt(n),temp = arr[j];
               arr[j] = arr[i];
               arr[i] = temp;
           }
           Arrays.sort(arr);
       }
       static long nCk(int n, int k) {
           return (modMult(fact(n),fastexp(modMult(fact(n-k),fact(k)),M-2)))%M;
       }
    
       static long fact (long n) {
           long fact =1;
           for(int i=1; i<=n; i++) {
               fact = modMult(fact,i)%M;
           }
           return fact%M;
       }
    
       static long modMult(long a,long b) {
           a=a%M; b=b%M;
           return ((a%M)*(b%M))%M;
       }
    
       static long fastexp(long x, long y){
           x=x&M;
           y=y%M;
           if(y==1) return x;
           long ans = fastexp(x,y/2);
           if(y%2 == 0) return modMult(ans,ans)%M;
           else return modMult(ans,modMult(ans,x))%M;
       }
   }