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.*;
  import java.math.*;
  import java.lang.*;
  
  public class MAXMEXCut {
      
      static int mod = 1000000007;
  
      private 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;
          }
      }
  
      public static void main(String[] args) throws Exception {
          FastReader scn = new FastReader();
          PrintWriter pw = new PrintWriter(System.out);
          int t = scn.nextInt();
          outer : while(t-->0){
              int n = scn.nextInt();
              String a = scn.nextLine();
              String b = scn.nextLine();
              long ans = 0L;
              int i = 0; 
              while(i < n){
              	if((a.charAt(i) == '0' && b.charAt(i) == '1') || (a.charAt(i) == '1' && b.charAt(i) == '0')){
              		ans += 2;
              		i++;
              	}else if(a.charAt(i) == '0' && b.charAt(i) == '0'){
              		if(i + 1 < n){
              			if(a.charAt(i+1) == '1' && b.charAt(i+1) == '1'){
              				ans += 2;
              				i += 2;
              			}else if(a.charAt(i+1) == '0' && b.charAt(i+1) == '0'){
              				ans += 1;
              				i++;
              			}else{
              				ans += 3;
              				i += 2;
              			}
              		}else{
              			ans += 1;
              			i++;
              		}
              	}else if(a.charAt(i) == '1' && b.charAt(i) == '1'){
              		if(i + 1 < n){
              			if(a.charAt(i+1) == '1' && b.charAt(i+1) == '1'){
              				ans += 0;
              				i++;
              			}else{
              				ans += 2;
              				i += 2;
              			}
              		}else{
              			ans += 0;
              			i++;
              		}
              	}
              }
              pw.println(ans);
          }
          pw.close();
      }
  
      public static class Pair{
          long x;
          long y;
  
          Pair(long x, long y){
              this.x = x;
              this.y = y;
          }
      }
  
      private static void sort(int[] arr) {
          List<Integer> list = new ArrayList<>();
          for (int i=0; i<arr.length; i++){
              list.add(arr[i]);
          }
          Collections.sort(list); // collections.sort uses nlogn in backend
          for (int i = 0; i < arr.length; i++){
              arr[i] = list.get(i);
          }
      }
  
      private static void sort(long[] arr) {
          List<Long> list = new ArrayList<>();
          for(int i=0; i<arr.length; i++){
              list.add(arr[i]);
          }
          Collections.sort(list); // collections.sort uses nlogn in backend
          for (int i = 0; i < arr.length; i++){
              arr[i] = list.get(i);
          }
      }
  
      private static int lowerBound(int[] arr, int x){
          int n = arr.length, si = 0, ei = n - 1;
          while(si <= ei){
              int mid = si + (ei - si)/2;
              if(arr[mid] == x){
                  if(mid-1 >= 0 && arr[mid-1] == arr[mid]){
                      ei = mid-1;
                  }else{
                      return mid;
                  }
              }else if(arr[mid] > x){
                  ei = mid - 1; 
              }else{
                  si = mid+1;
              }
          }
          return si; 
      }
  
      private static int upperBound(int[] arr, int x){
          int n = arr.length, si = 0, ei = n - 1;
          while(si <= ei){
              int mid = si + (ei - si)/2;
              if(arr[mid] == x){
                  if(mid+1 < n && arr[mid+1] == arr[mid]){
                      si = mid+1;
                  }else{
                      return mid + 1;
                  }
              }else if(arr[mid] > x){
                  ei = mid - 1; 
              }else{
                  si = mid+1;
              }
          }
          return si; 
      }
  
      private static int upperBound(ArrayList<Integer> list, int x){
          int n = list.size(), si = 0, ei = n - 1;
          while(si <= ei){
              int mid = si + (ei - si)/2;
              if(list.get(mid) == x){
                  if(mid+1 < n && list.get(mid+1) == list.get(mid)){
                      si = mid+1;
                  }else{
                      return mid + 1;
                  }
              }else if(list.get(mid) > x){
                  ei = mid - 1; 
              }else{
                  si = mid+1;
              }
          }
          return si; 
      }
  
      // (x^y)%p in O(logy)
      private static long power(long x, long y){
          long res = 1; 
          x = x % mod;
          while(y > 0){
              if ((y & 1) == 1){
                  res = (res * x) % mod;
              }
              y = y >> 1; 
              x = (x * x) % mod;
          }
          return res;
      }
  
      public static boolean nextPermutation(int[] arr) {
          if(arr == null || arr.length <= 1){
              return false;
          }
          
          int last = arr.length-2;
          while(last >= 0){
              if(arr[last] < arr[last+1]){
                  break;
              }
              last--;
          }
  
          if (last < 0){
                return false;
          }
          
          if(last >= 0){
              int nextGreater = arr.length-1;
              for(int i=arr.length-1; i>last; i--){
                  if(arr[i] > arr[last]){
                      nextGreater = i;
                      break;
                  }
              }
              swap(arr, last, nextGreater);
          }
          
          reverse(arr, last+1, arr.length-1);
          return true;
      }
      
      private static void swap(int[] arr, int i, int j){
          int temp = arr[i];
          arr[i] = arr[j];
          arr[j] = temp;
      }
      
      private static void reverse(int[] arr, int i, int j){
          while(i < j){
              swap(arr, i++, j--);
          }
      }
  
      // TC- O(logmax(a,b))
      private static int gcd(int a, int b) {
          if(a == 0){
              return b;
          }
          return gcd(b%a, a);
      }
  
      private static long gcd(long a, long b) {
          if(a == 0L){
              return b;
          }
          return gcd(b%a, a);
      }
  
      private static long lcm(long a, long b) {
          return a / gcd(a, b) * b;
      }
  
      // TC- O(logmax(a,b))
      private static int lcm(int a, int b) {
          return a / gcd(a, b) * b;
      }
  
      private static long inv(long x){
          return power(x, mod - 2);
      }
  
      private static long summation(long n){
          return (n * (n + 1L)) >> 1;
      } 
  }