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.*;
public class cforce {

	public static void main(String[] args) throws IOException
	{
		BufferedReader read = new BufferedReader(new InputStreamReader(System.in)); 
	     int t = Integer.parseInt(read.readLine());
	     while(t-- >0)
			{			    
	    	 int n=Integer.parseInt(read.readLine());
	    	 String s1=read.readLine();
	    	 String s2=read.readLine();
	    	 int s=0;
	    	 int a[]=new int[n];
			 for(int i=0;i<n;i++)
			 {
				s=Integer.parseInt(s1.substring(i,i+1))+Integer.parseInt(s2.substring(i,i+1));
				//System.out.println(s);
				if(s==0)
				a[i]=1;
				if(s==1)
					a[i]=2;
				if(s==2)
					a[i]=0;
			
			 }
			 //for(int i=0;i<n;i++)
				 //System.out.print(a[i]);
			 //System.out.println();
			 s=0;
			 for(int i=0;i<n;i++)
			 {
				// if()
				 if(i==n-1)
					 
					 {
					 s=s+a[n-1];
					 break;
					 }
				 else
				 {
					 if(a[i]==2)
						 s=s+2;
					 if(a[i]==1)
					 {
						 if(a[i+1]==0)
						 {
							 s=s+2;
							 i++;
						 }
						 else
						 {
							 s=s+1;
						 }
						continue;
					 }
					 if(a[i]==0)
					 {
						 if(a[i+1]==1)
						 {
							 s=s+2;
							 i++;
						 }
						 continue;
					 } 
				 }
				 
				 
			 }
			 System.out.println(s);
	}
	}

}
