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.*;
import java.lang.*;
public class MyClass {
    public static void main(String args[]) {
        
        Scanner sc = new Scanner(System.in);
        
        int t = sc.nextInt();
        sc.nextLine();
        
        while(t-- >0){
            
            int n = sc.nextInt();
            sc.nextLine();
            String str1 = sc.nextLine();
            String str2 = sc.nextLine();
            long res = 0l;
            
            for(int i=0 ; i<n ;){
                if((str1.charAt(i)=='0'&&str2.charAt(i)=='1') || (str2.charAt(i)=='0' && str1.charAt(i)=='1' )){
                    res+=2l;
                    i++;
                }else if((str1.charAt(i)=='1'&&str2.charAt(i)=='1') && i!=n-1 && (str1.charAt(i+1)=='0'&&str2.charAt(i+1)=='0')){
                    i+=2;
                    res+=2;
                }else if((str1.charAt(i)=='0'&&str2.charAt(i)=='0') &&  i!=n-1 && (str1.charAt(i+1)=='1'&&str2.charAt(i+1)=='1')){
                    i+=2;
                    res+=2;
                }else if((str1.charAt(i)=='0'&&str2.charAt(i)=='0')){
                    i+=1;
                    res+=1;
                }else{
                    i++;
                }
            }
            
            System.out.println(res);
            
            
             
            
        }
       
        
        
        
        
        
        
        
        
      
    }
}