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.FileNotFoundException;
  import java.io.FileReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.Arrays;
  import java.util.StringTokenizer;
  
  public class C {
  	public static void main(String[] args) {
  		FastScanner sc = new FastScanner();
  		int t = sc.nextInt();
  		for (int i = 0; i < t; i++) {
  			int n = sc.nextInt();
  			int[][] x = new int[2][n];
  			int[] cSum = new int[n];
  			char[] s1 = sc.nextToken().toCharArray();
  			char[] s2 = sc.nextToken().toCharArray();
  			for (int j = 0; j < n; j++) {
  				if (s1[j] == '0') {
  					x[0][j] = 0;
  				} else {
  					x[0][j] = 1;
  				}
  				if (s2[j] == '0') {
  					x[1][j] = 0;
  				} else {
  					x[1][j] = 1;
  				}
  				cSum[j] = x[0][j] + x[1][j];
  			}
  
  			int cc = 0;
  			int ans = 0;
  			while (cc < n) {
  				if (cc == n - 1) {
  					if (cSum[cc] == 0) {
  						ans += 1;
  						cc += 1;
  					} else if (cSum[cc] == 1) {
  						ans += 2;
  						cc += 1;
  					} else {
  						ans += 0;
  						cc += 1;
  					}
  				} else {
  
  					if (cSum[cc] == 0) { // 00
  						if (cSum[cc + 1] == 2) {
  							ans += 2;
  							cc += 2;
  						} else {
  							ans += 1;
  							cc += 1;
  						}
  					} else if (cSum[cc] == 1) {
  						ans += 2;
  						cc += 1;
  					} else {
  						if (cSum[cc + 1] == 0) {
  							ans += 2;
  							cc += 2;
  						} else {
  							ans += 0;
  							cc += 1;
  						}
  					}
  				}
  
  			}
  			System.out.println(ans);
  		}
  
  	}
  
  	public static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastScanner(String s) {
  			try {
  				br = new BufferedReader(new FileReader(s));
  			} catch (FileNotFoundException e) {
  				// TODO Auto-generated catch block
  				e.printStackTrace();
  			}
  		}
  
  		public FastScanner() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		String nextToken() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					// TODO Auto-generated catch block
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
  
  		long nextLong() {
  			return Long.parseLong(nextToken());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  	}
  
  }
  