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.lang.*;
import java.util.*;

public class a{
	public static Scanner scanner = new Scanner(System.in);
	
	public static void main(String[] args){
		 int t = scanner.nextInt();
		 scanner.nextLine();
		 while(t != 0){
			 t--;
			 function();
		 }
	}
	
	public static void function(){
		int n = scanner.nextInt();
		scanner.nextLine();
		String str1 = scanner.nextLine();
		String str2 = scanner.nextLine();
	//	HashMap<Integer,Integer> myhash = new HashMap<Integer,Integer>();
	//	myhash.put(0,0);
	//	myhash.put(1,0);
	//	myhash.put(2,0);
	    int ans = 0;
	  //  int track = 0;
	    boolean prevzero = false;
	    boolean prevone = false;
		for(int i = 0;i<n;i++){
			if(str1.charAt(i) != str2.charAt(i)){
				ans+=2;
				prevone = false;
				prevzero = false;
			}else if(str1.charAt(i) == '1'){
				  if(prevzero){
					  prevzero = false;
					  ans+=1;
				  }else{
					  prevone = true;
					  prevzero = false;
				  }
			}else{
				if(prevone){
					ans+=2;
					prevone = false;
				}else{
					 ans+=1;
					 prevzero = true;
				}  
			}
		}
		
		System.out.println(ans);

	}
	
}
