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 Main
{
    public static int count(String ans)
    {
        char prev='a';
       int c=0;
       int z=0;
        for(int i=0;i<ans.length();i++)
        {
            if(ans.charAt(i)==prev)
            {
                prev=ans.charAt(i);
            }
            else
            {
                c++;
                prev=ans.charAt(i);
            }
            
            
            if(ans.charAt(i)=='0')
            z++;
        }
       // System.out.println(z);
       int val=0;
       if(c%2==1)
       {
           if(ans.charAt(0)=='0')
           val=c/2+1;
           else
           val=c/2;
       }
       else
       val=c/2;
        return Math.max((val)*2,z);
    }
   public static int solve(String st1,String st2,int n)
    {
      int val=0;
     // int i=0;
      int j=0;
      String ans="";
     for(int i=0;i<n;i++)
     {
         
        if((st1.charAt(i)=='1' && st2.charAt(i)=='0') || (st1.charAt(i)=='0' && st2.charAt(i)=='1'))
        {
            val=val+count(ans);
           // System.out.println(ans);
            ans="";
            val=val+2;
        }
        else
        {
            ans=ans+st1.charAt(i);
        }
     }
     //System.out.println(ans+" "+count(ans));
    val+=count(ans);
      return val;
    }
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0)
		{
		    int n=sc.nextInt();
		  String st1=sc.next();
		  String st2=sc.next();
	//	System.out.println("op="+count("0111110"));
		   System.out.println(solve(st1,st2,n));
		}
	}
}
