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.*;
  import java.util.*;
   
   
  public class template {
      static class Reader {
          final private int BUFFER_SIZE = 1 << 16;
          private DataInputStream din;
          private byte[] buffer;
          private int bufferPointer, bytesRead;
   
          public Reader()
          {
              din = new DataInputStream(System.in);
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
   
          public Reader(String file_name) throws IOException
          {
              din = new DataInputStream(
                  new FileInputStream(file_name));
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
   
          public String readLine() throws IOException
          {
              byte[] buf = new byte[64]; // line length
              int cnt = 0, c;
              while ((c = read()) != -1) {
                  if (c == '\n') {
                      if (cnt != 0) {
                          break;
                      }
                      else {
                          continue;
                      }
                  }
                  buf[cnt++] = (byte)c;
              }
              return new String(buf, 0, cnt);
          }
   
          public int nextInt() throws IOException
          {
              int ret = 0;
              byte c = read();
              while (c <= ' ') {
                  c = read();
              }
              boolean neg = (c == '-');
              if (neg)
                  c = read();
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
   
              if (neg)
                  return -ret;
              return ret;
          }
   
          public long nextLong() throws IOException
          {
              long ret = 0;
              byte c = read();
              while (c <= ' ')
                  c = read();
              boolean neg = (c == '-');
              if (neg)
                  c = read();
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
              if (neg)
                  return -ret;
              return ret;
          }
   
          public double nextDouble() throws IOException
          {
              double ret = 0, div = 1;
              byte c = read();
              while (c <= ' ')
                  c = read();
              boolean neg = (c == '-');
              if (neg)
                  c = read();
   
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
   
              if (c == '.') {
                  while ((c = read()) >= '0' && c <= '9') {
                      ret += (c - '0') / (div *= 10);
                  }
              }
   
              if (neg)
                  return -ret;
              return ret;
          }
   
          private void fillBuffer() throws IOException
          {
              bytesRead = din.read(buffer, bufferPointer = 0,
                                   BUFFER_SIZE);
              if (bytesRead == -1)
                  buffer[0] = -1;
          }
   
          private byte read() throws IOException
          {
              if (bufferPointer == bytesRead)
                  fillBuffer();
              return buffer[bufferPointer++];
          }
   
          public void close() throws IOException
          {
              if (din == null)
                  return;
              din.close();
          }
      }
   
      public static void main(String[] args)
          throws IOException
      {
          Scanner s = new Scanner(System.in);
          int t = s.nextInt();
          while (t-->0) {
          int flag = 0;
          long n = s.nextInt();
          String s1 = s.next();
          String s2 = s.next();
          int ans =0;
          for(int i=0;i<s1.length();i++)
          {
               if(s1.charAt(i)=='1' && s2.charAt(i)=='1')
          {
              while(i<s1.length() && (s1.charAt(i)=='1' && s2.charAt(i)=='1'))
              {
                  i++;
              }
              if(!(i==s1.length()))
              {
                  ans+=2;
              }
          }
          else
          {
              if(s1.charAt(i)=='0' && s2.charAt(i)=='0')
              {
                  if(i+1<s1.length() && s1.charAt(i+1)=='1' && s2.charAt(i+1)=='1')
                  {
                      ans+=2;
                     i++;
                  }
                  else
                      ans++;
              }
              else
              {
                  ans=ans+2;
              }
          }
          }
        System.out.println(ans);
          
        
      }
      }}