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.*;

public class A
{
    static void sort(int[] A){

        int n = A.length;
        Random rnd = new Random();
        for(int i=0; i<n; ++i){
            int tmp = A[i];
            int randomPos = i + rnd.nextInt(n-i);
            A[i] = A[randomPos];
            A[randomPos] = tmp;
        }
        Arrays.sort(A);
    }

    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        long mod=1000000007l;
        int cases=sc.nextInt();

        while(cases>0)
        {
            cases--;
            int n=sc.nextInt();
            String s1=sc.next();
            String s2=sc.next();

            boolean one=false;
            boolean zero=false;
            int ans=0;

            for(int j=0;j<n;j++)
            {
                if(s1.charAt(j)=='0'||s2.charAt(j)=='0')
                {
                    zero=true;
                }
                if(s1.charAt(j)=='1'||s2.charAt(j)=='1')
                {
                    one=true;
                }

                if(one&&zero)
                {
                    ans=ans+2;
                    one=false;
                    zero=false;
                }
                else if(zero==true&&one==false)
                {
                    if(j!=n-1&& s1.charAt(j+1)=='1'&&s2.charAt(j+1)=='1')
                    {
                        j++;
                        ans=ans+2;
                        zero=false;
                        one=false;
                    }
                    else
                    {
                        ans++;
                        zero=false;
                    }
                }

            }

            System.out.println(ans);


        }

    }
}