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.util.Scanner;

public class Main3 {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = Integer.valueOf(in.nextLine());
        for (int i = 0; i < t; i++) {
            int n = Integer.valueOf(in.nextLine());
            String s1 = in.nextLine();
            String s2 = in.nextLine();
            calc(s1, s2, n);
        }
    }

    private static void calc(String s1, String s2, int n) {
        int result = 0;
        int i = 0;
        while(i < n) {
            if(s1.charAt(i) != s2.charAt(i)) {
                result += 2;
                i++;
            }
            else {
                if (s1.charAt(i) == '0') {
                    if (i < n - 1) {
                        if(s1.charAt(i + 1) == '1' && s2.charAt(i + 1) == '1') {
                            result++;
                            i++;
                        }
                        result++;
                        i++;
                    }
                    else {
                        result++;
                        i++;
                    }
                }
                else {
                    if (i < n - 1 && s1.charAt(i + 1) == '0' && s2.charAt(i + 1) == '0') {
                        result += 2;
                        i+= 2;
                    }
                    else {
                        i++;
                    }
                }
            }
        }
        System.out.println(result);
    }
}