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.*;
import java.util.*;

public class C {
	 
	public static void main(String[] args) {
		
		FastReader f = new FastReader();
		StringBuffer sb=new StringBuffer();
		
	   int test=f.nextInt();
	   while(test-->0)
	   {
		   int n=f.nextInt();
		   char str1[]=f.next().toCharArray();
		   char str2[]=f.next().toCharArray();
		   
		   long ans=0;
		   boolean selected = false;
		    
		    for (int i = 0; i < n; i++)
		    {
		        if (str2[i] != str1[i])
		        {
		            if (selected)
		            {
		                if (str1[i-1] == '0') 
		                	ans += 1;
		                selected = false;
		                ans += 2;
		            }
		            else
		                ans += 2;
		        }
		        else 
		        {
		            if (selected)
		            {
		                if (str1[i-1] == str1[i]) 
		                {
		                    if (str1[i] == '0') 
		                    	ans++;
		                }
		                else {
		                    ans += 2;
		                    selected = false;
		                }
		            }
		            else 
		            {
		                selected = true;
		            }
		        }
		    }
		    if (selected)
		    {
		        if (str1[n-1] == '0') ans++;
		    }
		    sb.append(ans+"\n");
	   }
	   System.out.println(sb);
	}
	
	
	static class FastReader 
	{ 
	    BufferedReader br; 
	    StringTokenizer st; 
 
	    public FastReader() {
	    	br = new BufferedReader(new
	                 InputStreamReader(System.in)); 
	    }
	    String next() { 
	        while (st == null || !st.hasMoreElements()) { 
	            try{ 
	                st = new StringTokenizer(br.readLine()); 
	            } 
	            catch (IOException  e) { 
	                e.printStackTrace(); 
	            } 
	        } 
	        return st.nextToken(); 
	    } 
	    int nextInt() { 
	        return Integer.parseInt(next()); 
	    } 
	    long nextLong() { 
	        return Long.parseLong(next()); 
	    } 
	    double nextDouble() { 
	        return Double.parseDouble(next()); 
	    } 
	    String nextLine() { 
	        String str = ""; 
	        try{ 
	            str = br.readLine(); 
	        } 
	        catch (IOException e) { 
	            e.printStackTrace(); 
	        } 
	        return str; 
	    } 
	} 
}