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 MaxMexCut {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = Integer.parseInt(sc.nextLine());
        for(int i=0; i<t; i++) {
            int n = Integer.parseInt(sc.nextLine());
            char[][] biTable = new char[2][n];
            String s1 = sc.nextLine();
            String s2 = sc.nextLine();

            for(int j=0; j<n; j++) {
                biTable[0][j] = s1.charAt(j);
                biTable[1][j] = s2.charAt(j);
            }
            System.out.println(maxMex(biTable));
        }
    }

    public static int maxMex(char[][] biTable) {
        int maxMex = 0;
        boolean prevZero = false;
        boolean prevOne = false;

        for(int i=0; i< biTable[0].length; i++) {
            if(biTable[0][i] != biTable[1][i]) {
                maxMex += 2;
                prevZero = prevOne = false;
            } else {
                boolean currZero = biTable[0][i] == '0';
                boolean currOne = biTable[0][i] == '1';

                // Take the current column into account
                if(currZero) {
                    maxMex += 1;
                }
                if((prevZero && currOne) || (prevOne && currZero)) {
                    maxMex += 1;
                    prevZero = false; prevOne = false;
                    continue;
                }
                // Set prev zero or prev one
                prevOne = currOne;
                prevZero = currZero;
            }

        }

        return maxMex;
    }
}
