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.ByteArrayInputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.util.Arrays;
  import java.util.InputMismatchException;
  
  public class C {
  	private static PrintWriter out = new PrintWriter(System.out);
  
  	public static void solve() {
  		In in = new In();
  		int tests = in.ni();
  		while (tests-- > 0) {
  			int n = in.ni();
  			char[] s = in.ns().toCharArray();
  			char[] t = in.ns().toCharArray();
  			long ans = 0;
  			for (int i = 0; i < n; i++) {
  				if (s[i] != t[i]) {
  					ans += 2;
  				} else if (s[i] == '0') {
  					if (i + 1 < n) {
  						if (s[i + 1] != s[i] && t[i + 1] != s[i]) {
  							ans += 2;
  							i++;
  						} else {
  							ans++;
  						}
  					} else {
  						ans++;
  					}
  				} else { // s[i] == t[i] == '1'
  					if (i + 1 < n && (s[i + 1] != s[i] || t[i + 1] != s[i])) {
  						ans += 2;
  						i++;
  					}
  				}
  			}
  			out.println(ans);
  
  		}
  	}
  
  	public static void main(String[] args) throws IOException {
  		// long start = System.nanoTime();
  		solve();
  		// System.out.println("Elapsed: " + (System.nanoTime() - start) / 1000000 +
  		// "ns");
  		out.flush();
  	}
  
  	@SuppressWarnings("unused")
  	private static class In {
  		final private static int BUFFER_SIZE = 1024;
  		private byte[] buf;
  		private InputStream is;
  		private int buflen;
  		private int bufptr;
  
  		public In() {
  			is = System.in;
  			buf = new byte[BUFFER_SIZE];
  			buflen = bufptr = 0;
  		}
  
  		public In(String input) {
  			is = new ByteArrayInputStream(input.getBytes());
  			buf = new byte[BUFFER_SIZE];
  			buflen = bufptr = 0;
  		}
  
  		public int readByte() {
  			if (bufptr >= buflen) {
  				try {
  					buflen = is.read(buf);
  				} catch (IOException ioe) {
  					throw new InputMismatchException();
  				}
  				bufptr = 0;
  			}
  			if (buflen <= 0)
  				return -1;
  			return buf[bufptr++];
  		}
  
  		public boolean isSpaceChar(int c) {
  			return !(c >= 33 && c <= 126);
  		}
  
  		public int skip() {
  			int b;
  			while ((b = readByte()) != -1 && isSpaceChar(b))
  				;
  			return b;
  		}
  
  		/* Next character */
  		public char nc() {
  			return (char) skip();
  		}
  
  		/* Next double */
  		public double nd() {
  			return Double.parseDouble(ns());
  		}
  
  		/* Next string */
  		public String ns() {
  			final StringBuilder sb = new StringBuilder();
  			int b = skip();
  			while (!isSpaceChar(b)) {
  				sb.appendCodePoint(b);
  				b = readByte();
  			}
  			return sb.toString();
  		}
  
  		/* Next integer */
  		public int ni() {
  			boolean minus = false;
  			int num = 0;
  			int b;
  			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
  				;
  			if (b == '-') {
  				minus = true;
  				b = readByte();
  			}
  			while (b >= '0' && b <= '9') {
  				num = num * 10 + (b - '0');
  				b = readByte();
  			}
  			return minus ? -num : num;
  		}
  
  		/* Next long */
  		public long nl() {
  			boolean minus = false;
  			long num = 0;
  			int b;
  			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
  				;
  			if (b == '-') {
  				minus = true;
  				b = readByte();
  			}
  			while (b >= '0' && b <= '9') {
  				num = num * 10 + (b - '0');
  				b = readByte();
  			}
  			return minus ? -num : num;
  		}
  
  		/* Next integer 1D array */
  		public Integer[] nia(int n) {
  			final Integer[] arr = new Integer[n];
  			for (int i = 0; i < n; i++)
  				arr[i] = ni();
  			return arr;
  		}
  
  		/* Next long 1D array */
  		public long[] nla(int n) {
  			final long[] arr = new long[n];
  			for (int i = 0; i < n; i++)
  				arr[i] = nl();
  			return arr;
  		}
  
  		/* Next string 1D array */
  		public String[] nsa(int n) {
  			final String[] arr = new String[n];
  			for (int i = 0; i < n; i++)
  				arr[i] = ns();
  			return arr;
  		}
  
  		/* Next char 1D array */
  		public char[] nca(int n) {
  			final char[] arr = new char[n];
  			for (int i = 0; i < n; i++)
  				arr[i] = nc();
  			return arr;
  		}
  
  		/* Next double 1D array */
  		public double[] nda(int n) {
  			final double[] arr = new double[n];
  			for (int i = 0; i < n; i++)
  				arr[i] = nc();
  			return arr;
  		}
  
  		/* Next integer matrix */
  		public int[][] nim(int n, int m) {
  			final int[][] arr = new int[n][m];
  			for (int i = 0; i < n; i++)
  				for (int j = 0; j < m; j++)
  					arr[i][j] = ni();
  			return arr;
  		}
  
  		/* Next long matrix */
  		public long[][] nlm(int n, int m) {
  			final long[][] arr = new long[n][m];
  			for (int i = 0; i < n; i++)
  				for (int j = 0; j < m; j++)
  					arr[i][j] = nl();
  			return arr;
  		}
  
  		/* Next string matrix */
  		public String[][] nsm(int n, int m) {
  			final String[][] arr = new String[n][m];
  			for (int i = 0; i < n; i++)
  				for (int j = 0; j < m; j++)
  					arr[i][j] = ns();
  			return arr;
  		}
  
  		/* Next char matrix */
  		public char[][] ncm(int n, int m) {
  			final char[][] arr = new char[n][m];
  			for (int i = 0; i < n; i++)
  				for (int j = 0; j < m; j++)
  					arr[i][j] = nc();
  			return arr;
  		}
  
  		/* Next double matrix */
  		public double[][] ndm(int n, int m) {
  			final double[][] arr = new double[n][m];
  			for (int i = 0; i < n; i++)
  				for (int j = 0; j < m; j++)
  					arr[i][j] = nd();
  			return arr;
  		}
  
  		public static void log(Object... o) {
  			System.out.println(Arrays.deepToString(o));
  		}
  	}
  
  }
  