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.
import java.io.BufferedReader;
import java.util.StringTokenizer;
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.*;
public class C {


   

    public static void main(String[] args) {
        FastReader fr = new FastReader();
        PrintWriter out = new PrintWriter(System.out, true);
        int cases = fr.nextInt();
        for(int c = 0; c < cases; c++) {
            int size = fr.nextInt();
            String astring = fr.next();
            String bstring = fr.next();

            int max = 0;
            int i = 0;
            while(i < size) {
                char achar = astring.charAt(i);
                char bchar = bstring.charAt(i);
                if(achar == '1' && bchar == '0' || bchar == '1' && achar == '0') {
                    i++;
                    max += 2;
                
                }

                else if(achar == '0' && bchar == '0') {
                    if(i < astring.length() - 1) {
                        if(astring.charAt(i+1) == '1' && bstring.charAt(i+1) == '1') {
                            max += 2;
                            i += 2;
                        } else {
                            i++;
                            max += 1;
                        }
                    } else {
                        i++;
                        max += 1;
                        
                    }
                }

                else if(achar == '1' && bchar == '1') {
                    while(i < size && achar == '1' && bchar == '1') {
                        achar = astring.charAt(i);
                        bchar = bstring.charAt(i);
                        i++;
                    }
                    if(i < size) {
                        max += 2;
                    }
                   
                }

            }

            out.write(max + "\n");
        }
        out.close();
    }

    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            this.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;
        }
    }
}
