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.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.OutputStream;
  import java.io.PrintWriter;
  import java.io.BufferedWriter;
  import java.io.Writer;
  import java.io.OutputStreamWriter;
  import java.util.InputMismatchException;
  import java.io.IOException;
  import java.io.InputStream;
  
  /**
   * Built using CHelper plug-in
   * Actual solution is at the top
   */
  public class Main {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          InputReader in = new InputReader(inputStream);
          OutputWriter out = new OutputWriter(outputStream);
          CMAXMEXCut solver = new CMAXMEXCut();
          solver.solve(1, in, out);
          out.close();
      }
  
      static class CMAXMEXCut {
          public void solve(int testNumber, InputReader in, OutputWriter out) {
              int tc = in.nextInt();
              for (; tc > 0; tc--) {
                  int size = in.nextInt();
                  String s = in.nextString();
                  String s2 = in.nextString();
                  // check number of separated zeros
                  int prev00 = -10;
                  long sum = 0;
                  long prev11 = -10;
                  for (int i = 0; i < s.length(); i++) {
                      char c = s.charAt(i);
                      char c1 = s2.charAt(i);
                      if (c + c1 == '0' + '0') {
                          if (prev11 == i - 1) {
                              //join
                              sum += 2;
                          } else {
                              sum++;
                              prev00 = i;
                          }
                      } else if (c + c1 == '1' + '1') {
                          if (prev00 == i - 1) {
                              // use it
                              sum += 1;
                          } else {
                              prev11 = i;
                          }
                      } else if (c + c1 == '0' + '1') {
                          sum += 2;
                      }
                  }
  
                  out.println(sum);
              }
  
              //cases 0 1 = 2
              // cases 0 0 = 1
              // cases 1 1 = 0
              // we want to avoid 1 1s
              // match 11 with a 0 0 to get a 2 2
              // if you have a 1 1 with a 0 0 match, otherwise don't bother
              // should we always match 0 0 with the next one? yeah
          }
  
      }
  
      static class InputReader {
          private InputStream stream;
          private byte[] buf = new byte[1024];
          private int curChar;
          private int numChars;
          private InputReader.SpaceCharFilter filter;
  
          public InputReader(InputStream stream) {
              this.stream = stream;
          }
  
          public int read() {
              if (numChars == -1) {
                  throw new InputMismatchException();
              }
              if (curChar >= numChars) {
                  curChar = 0;
                  try {
                      numChars = stream.read(buf);
                  } catch (IOException e) {
                      throw new InputMismatchException();
                  }
                  if (numChars <= 0) {
                      return -1;
                  }
              }
              return buf[curChar++];
          }
  
          public int nextInt() {
              int c = read();
              while (isSpaceChar(c)) {
                  c = read();
              }
              int sgn = 1;
              if (c == '-') {
                  sgn = -1;
                  c = read();
              }
              int res = 0;
              do {
                  if (c < '0' || c > '9') {
                      throw new InputMismatchException();
                  }
                  res *= 10;
                  res += c - '0';
                  c = read();
              } while (!isSpaceChar(c));
              return res * sgn;
          }
  
          public String nextString() {
              int c = read();
              while (isSpaceChar(c)) {
                  c = read();
              }
              StringBuilder res = new StringBuilder();
              do {
                  if (Character.isValidCodePoint(c)) {
                      res.appendCodePoint(c);
                  }
                  c = read();
              } while (!isSpaceChar(c));
              return res.toString();
          }
  
          public boolean isSpaceChar(int c) {
              if (filter != null) {
                  return filter.isSpaceChar(c);
              }
              return isWhitespace(c);
          }
  
          public static boolean isWhitespace(int c) {
              return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
          }
  
          public interface SpaceCharFilter {
              public boolean isSpaceChar(int ch);
  
          }
  
      }
  
      static class OutputWriter {
          private final PrintWriter writer;
  
          public OutputWriter(OutputStream outputStream) {
              writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
          }
  
          public OutputWriter(Writer writer) {
              this.writer = new PrintWriter(writer);
          }
  
          public void close() {
              writer.close();
          }
  
          public void println(long i) {
              writer.println(i);
          }
  
      }
  }
  
  