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.PrintWriter;
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner obj = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int length = obj.nextInt();
        while (length-- != 0) {
            int n = obj.nextInt();
            int[][] num = new int[2][n];
            for (int i = 0; i < 2; i++) {
                String s = obj.next();
                for (int j = 0; j < n; j++) {
                    num[i][j] = s.charAt(j) - '0';
                }
            }
            int count = 0;
            int[] check = new int[n];
            for (int i = 0; i < n - 1; i++) {
                if (check[i] == 1) continue;
                if ((num[0][i] == 1 && num[1][i] == 0) || (num[0][i] == 0 && num[1][i] == 1)) {
                    count += 2;
                } else if (num[0][i] == 1 && num[1][i] == 1) {
                    if ((num[0][i + 1] == 0 || num[1][i + 1] == 0)) {
                        count += 2;
                        check[i + 1] = 1;
                    }
                } else if (num[0][i] == 0 && num[1][i] == 0) {
                    if (num[0][i + 1] == 1 && num[1][i + 1] == 1) {
                        count += 2;
                        check[i + 1] = 1;
                    } else count += 1;
                }
                check[i] = 1;
            }
            if(check[n-1]==0)
            {
                if ((num[0][n-1] == 1 && num[1][n-1] == 0) || (num[0][n-1] == 0 && num[1][n-1] == 1)) {
                    count += 2;
                }
                else if (num[0][n-1] == 0 && num[1][n-1] == 0)
                {
                    count+=1;
                }
            }
            out.println(count);
        }
        out.flush();
    }
}