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.*;
import java.util.*;

public class MaxMex{ // start class

//_______________________________main method__________________________________

    public static void main(String args[]) throws IOException { // start method

        BufferedReader fin = new BufferedReader(new InputStreamReader(System.in));

        int T = Integer.parseInt(fin.readLine());
        for (int t = 0; t < T; t++) {
            int N = Integer.parseInt(fin.readLine());
            String top = fin.readLine();
            String bot = fin.readLine();
            int sum = 0;
            for (int i = 0; i < N; i++) {
                int a = Integer.parseInt(top.substring(i, i+1)), b = Integer.parseInt(bot.substring(i, i+1));
                if (a != b) sum += 2; // dif
                else { // same
                    if (i < N-1) {
                        int c = Integer.parseInt(top.substring(i+1, i+2)), d = Integer.parseInt(bot.substring(i+1, i+2));
                        if (c == d) { // same same
                            if (a == 0 && c == 0) sum += 1; // 00 00
                            else if (a != c) {sum += 2; i++; } // 00 11 or 11 00
                        }
                        else { // same dif
                            if (a == 0) sum += 1; // 00 dif
                        }
                    }
                    else {
                        if (a == 0) sum += 1; // 00
                    }
                }
            }
            System.out.println(sum);
        }
    }
}
