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 test {
  
      static long mod = 1000000007;
      static long inf = Long.MAX_VALUE;
      static final int maxSizeSeive = 1000001;
      static Vector<Boolean>isprime = new Vector<>(maxSizeSeive);
      static Vector<Integer>prime = new Vector<>();
      static Vector<Integer>SPF = new Vector<>(maxSizeSeive);
  
      public static void solve(int n, char[] s1, char[] s2) {
          boolean visited[] = new boolean[n];
          int[] group2 = new int[n];
          int numberOfGrp2 = 0;
          int[] group1 = new int[n];
          int[] group0 = new int[n];
          for(int i = 0; i < n; i++) {
              if(s1[i] == '1' && s2[i] == '0') {
                  group2[i] = 1;
                  numberOfGrp2 += 1;
                  visited[i] = true;
              }
              else if(s2[i] == '1' && s1[i] == '0') {
                  group2[i] = 1;
                  numberOfGrp2 += 1;
                  visited[i] = true;
              }
              else if(s2[i] == '1' && s1[i] == '1') {
                  group1[i] = 1;
              }
              else if(s2[i] == '0' && s1[i] == '0') {
                  group0[i] = 1;
              }
          }
          //System.out.println("test");
          if(numberOfGrp2 == n) {
              System.out.println((n*2));
              return;
          }
          //System.out.println("number of grp 2 : " + numberOfGrp2);
  
          /*for(int i  : group0) {
              System.out.print(i);
          }
          System.out.println();
          for(int i : group1) {
              System.out.print(i);
          }
          System.out.println();*/
          /*System.out.println("grping");
  
          int[] grouping = new int[n];
          for(int i = 0; i < n; i++) {
              grouping[i] = group0[i] + group1[i];
              System.out.print(grouping[i]);
          }
          System.out.println();*/
          
          for(int i = 1; i < n; i++) {
              if(!visited[i] && !visited[i - 1]) {
                  if(group0[i] == 1 && group1[i - 1] == 1) {
                      visited[i] = true;
                      visited[i - 1] = true;
                      numberOfGrp2 += 1;
                  }
                  else  if(group1[i] == 1 && group0[i - 1] == 1) {
                      visited[i] = true;
                      visited[i - 1] = true;
                      numberOfGrp2 += 1;
                  }
              }
          }
  
          //System.out.println("number of grp 2 : " + numberOfGrp2);
          long sum = numberOfGrp2 * 2;
          for(int i = 0; i < n; i++) {
              if(!visited[i]) {
                  if(group0[i] == 1) {
                      sum += 1;
                  }
              }
          }
  
          System.out.println(sum);
      }
      
  /*
  input:
  4
  7
  0101000
  1101100
  5
  01100
  10101
  2
  01
  01
  6
  000000
  111111
  
  8
  10011100
  01011100
  
  output:
  8
  8
  2
  12
  
  */
  
      public static void main(String[] args) {
          FastReader sc = new FastReader();
          int t = sc.nextInt();
          while(t-- > 0) {
              int n = sc.nextInt();
              char[] s1 = sc.nextLine().toCharArray();
              char[] s2 = sc.nextLine().toCharArray();
              // solve
              solve(n, s1, s2);
          }
      }
  
      public static ArrayList<Integer> seivePrime(int n) {
          ArrayList<Integer> primes = new ArrayList<>();
          for (int i = 0; i < maxSizeSeive; i++){
              isprime.add(true);
              SPF.add(2);
          }
  
          manipulated_seive(n);
  
          for (int i=0; i<prime.size() && prime.get(i) <= n; i++) {
              primes.add(prime.get(i));
          }
          return primes;
      }
  
      public static void manipulated_seive(int N) {
          isprime.set(0, false);
          isprime.set(1, false);
           
          for (int i=2; i<N ; i++) {
              if (isprime.get(i)) {
                  prime.add(i);
                  SPF.set(i,i);
              }
  
              for (int j=0;j < prime.size() && i*prime.get(j) < N && prime.get(j) <= SPF.get(i);j++) {
                  isprime.set(i*prime.get(j),false);
                  SPF.set(i*prime.get(j),prime.get(j));
              }
          }
      }    
       
  	public static void sort(long[] a) {
  		ArrayList<Long> l=new ArrayList<>();
  		for (long i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
  
      public static int maxIndex(long[] arr) {
          long max = Long.MIN_VALUE;
          int index = 0;
          int n = arr.length;
          for(int i = 0; i < n; i++) {
              if(arr[i] > max) {
                  index = i;
                  max = arr[i];
              }
          }
          return index;
      }
  
      public static int minIndex(long[] arr) {
          long min = Long.MAX_VALUE;
          int index = 0;
          int n = arr.length;
          for(int i = 0; i < n; i++) {
              if(arr[i] < min) {
                  index = i;
                  min = arr[i];
              }
          }
          return index;
      }
  
      public static void debugArray(long[] arr) {
          for(int i = 0; i < arr.length; i++) {
              System.out.print(arr[i] + " ");
          }
          System.out.println();
      } 
  
      public static void debugArrayList(ArrayList<Long> arr) {
          for(Long i : arr) {
              System.out.print(i + " ");
          }
          System.out.println();
      }
  
      public static long gcd(long a, long b) {
          if(b == 0) {
              return a;
          }
          else {
              return gcd(b, a % b);
          }
      }
   
      public static long LCM(long a, long b) {
          return Math.abs(a*b) / gcd(a, b);
      }
  
      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();
          }
  
          String nextLine() {
              String str="";
              try {
                  str=br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          long[] readArray(int n) {
              long[] a = new long[n];
              for(int i = 0; i < n ; i++) {
                  a[i] = nextLong();
              }
              return a;
          }
      }
  }