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.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
   
  public class C {
  public static void main(String[] args) {
  		
  		FastScanner sc = new FastScanner();
  		int t = sc.nextInt();
  		for(int tt = 0; tt < t ; tt++) {
  			int n = sc.nextInt();
  			char a[][] = new char[2][n];
  			
  			a[0] = sc.next().toCharArray();
  			a[1] = sc.next().toCharArray();
  			
  			int ans = 0;
  	outer:		for(int i = 0 ; i < n ; i++) {
  				if(i < n-1) {
  					if((a[0][i] == '0' && a[1][i] == '0' && a[0][i+1] == '1' && a[1][i+1] == '1') || (a[0][i] == '1' && a[1][i] == '1' && a[0][i+1] == '0' && a[1][i+1] == '0')) {
  						ans += 2;
  						i++;
  						continue outer;
  					}
  				}
  				if(a[0][i] == '0' && a[1][i] == '0')ans += 1;
  				if((a[0][i] == '1' && a[1][i] == '0') || (a[0][i] == '0' && a[1][i] == '1')) {
  					ans += 2;
  				}
  			}
  			System.out.println(ans);
  		}
  	}
  	
  	
  	public static void sort(int[] a) {
  		ArrayList<Integer> l=new ArrayList<>();
  		for (int i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
  
    
      public static long gcd(long a, long b) {
          return b == 0 ? a : gcd(b, a % b);
      }
   
      public static int gcd(int a, int b) {
          return b == 0 ? a : gcd(b, a % b);
      }
      public static long pow(long base, long exp, long MOD) {
          base %= MOD;
          long ret = 1;
          while (exp > 0) {
              if ((exp & 1) == 1) ret = ret * base % MOD;
              base = base * base % MOD;
              exp >>= 1;
          }
          return ret;
      }
   
   
      public static long max(long... ar) {
          long ret = ar[0];
          for (long itr : ar) ret = Math.max(ret, itr);
          return ret;
      }
   
      public static int max(int... ar) {
          int ret = ar[0];
          for (int itr : ar) ret = Math.max(ret, itr);
          return ret;
      }
   
      public static long min(long... ar) {
          long ret = ar[0];
          for (long itr : ar) ret = Math.min(ret, itr);
          return ret;
      }
   
      public static int min(int... ar) {
          int ret = ar[0];
          for (int itr : ar) ret = Math.min(ret, itr);
          return ret;
      }
   
   
      public static long sum(long... ar) {
          long sum = 0;
          for (long itr : ar) sum += itr;
          return sum;
      }
   
      public static long sum(int... ar) {
          long sum = 0;
          for (int itr : ar) sum += itr;
          return sum;
      }
  	
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		
  		public String next() {
  			while (!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());
  		}
  		
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  	}
   
  }
  
  
   
   
   
   
   
   
   
   
  