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.Scanner;

public class JavaApplication68 {
    public static void main(String[] args) {
        Scanner sc = new Scanner (System.in);
        int t = sc.nextInt();
        while(t-->0) {
        
        //System.out.println("111");
     
        int n = sc.nextInt();
        //System.out.println("////");
        int [] arr = new int [n];
        
        //System.out.println("aa");
        
        String str1 = sc.next();
        String str2 = sc.next();
        for(int i=0; i<n; i++){
            arr [i] =  Integer.parseInt(String.valueOf(str1.charAt(i ))) +Integer.parseInt(String.valueOf(str2.charAt(i )));
             
        }
        //for(int x: arr){System.out.println(x);}
        
        int count = 0, prev = 1;
        for(int i =0; i< n ; i++){
            
            if(arr[i] == 1)     {if(prev == 0){count += 3; prev = 1;} else {count += 2; prev = 1;}  }
            
            else if (arr[i] == 2)          { if (prev == 0){count += 2; prev=1; } else { prev= 2;  }}
            
            else           { if(i== n-1){ count++;  if(prev == 0 || prev == 2) count++;  break;}     if(prev==1) prev =0; else if (prev == 0) count++; else {count +=2; prev =1;}  }
            
        }
        System.out.println(count);
        
        }
       
    }
}