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.StringTokenizer;
  import java.util.*;
  
  public class MaxMex {
      static final Random random = new Random();
  
      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;
          }
  
          int[] readArray(int n, int size) {
              int[] a = new int[size];
              for (int i = 0; i < n; i++) {
                  a[i] = this.nextInt();
              }
  
              return a;
          }
  
          long[] readLongArray(int n, int size) {
              long[] a = new long[size];
              for (int i = 0; i < n; i++) {
                  a[i] = this.nextLong();
              }
              return a;
          }
      }
  
      static void ruffleSort(long[] a) {
          int n = a.length;//shuffle, then sort
          for (int i = 0; i < n; i++) {
              int oi = random.nextInt(n);
              long temp = a[oi];
              a[oi] = a[i];
              a[i] = temp;
          }
          Arrays.sort(a);
      }
  
      static void ruffleSort(int[] a) {
          int n = a.length;//shuffle, then sort
          for (int i = 0; i < n; i++) {
              int oi = random.nextInt(n), temp = a[oi];
              a[oi] = a[i];
              a[i] = temp;
          }
          Arrays.sort(a);
      }
  
      public static void main(String[] args) {
          FastReader fs = new FastReader();
          int t = fs.nextInt();
  
          for (int z = 0; z < t; z++) {
              int n = fs.nextInt();
              char[] row1 = fs.next().toCharArray();
              char[] row2 = fs.next().toCharArray();
  
              int last_0_0 = -1;
              int last_1_1 = -1;
              long ans =0;
              for(int i = 0; i < n; i++) {
                  int sum = (row1[i] - '0') + (row2[i] - '0');
                  if(sum == 1) {
                      ans += 2;
                      last_0_0 = -1;
                      last_1_1 = -1;
                  }else if(sum == 2) {
                      if(last_0_0 != -1) {
                          ans += 1;
                          last_0_0 = -1;
                      }else {
                          last_1_1 = i;
                      }
                  }else {
                      if(last_1_1 != -1) {
                          ans += 2;
                          last_1_1 = -1;
                      }else {
                          ans += 1;
                          last_0_0 = i;
                      }
                  }
              }
              System.out.println(ans);
          }
      }
  }