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 class251 {
public static void main(String arg[])
{
     Scanner sc=new Scanner(System.in);
     int t=sc.nextInt();
     while(t-->0)
     {
    	int n=sc.nextInt();
    	String s1,s2;
    	s1=sc.next();
    	s2=sc.next();
    	char ch1[]=s1.toCharArray();
    	char ch2[]=s2.toCharArray();
    	int i,j,sum=0,c=0;
    	for(i=0;i<n;i++)
    	{
    		if(ch1[i]=='1' && ch2[i]=='0' || ch1[i]=='0' && ch2[i]=='1' )
    		{
    			sum+=2;
    		}
    		else if(ch1[i]=='0' && ch2[i]=='0' )
    		{
    			if(i<n-1 && ch1[i+1]=='1' && ch2[i+1]=='1' )
    			{
    				sum+=2;
    				i++;
    			}
    			else
    			{
    				sum++;
    			}
    		}
    		else
    		{
    			if( i<n-1 && ch1[i+1]=='0' && ch2[i+1]=='0')
    			{
    				sum+=2;
    				i++;
    			}
    		
    		}
    	}
    	System.out.println(sum);
     }
}
}
