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.*;

public class CFR16Q3{
	public static void main(String[] args)
	{
		Scanner sc = new Scanner(System.in);
	    
		int t = sc.nextInt();
		
	    while(t-- > 0)
		{
			int n = sc.nextInt();
			
			sc.nextLine();
			String s1 = sc.nextLine();
			String s2 = sc.nextLine();
			
			if(!s1.contains("0") && !s2.contains("0"))
			{
				System.out.println(0);
			}
			else if(!s1.contains("2") && !s2.contains("2"))
			{
				int res = 0;
				for(int i = 0; i < s1.length(); i++)
				{
					if(s1.charAt(i) == '0' && s2.charAt(i) == '0')
					{
						if(i < s1.length() - 1 && s1.charAt(i+1) == '1' && s2.charAt(i+1) == '1')
						{
							res += 2;
							i++;
						}
						else{
							res++;
						}
					}
					else if(s1.charAt(i) == '1' && s2.charAt(i) == '1')
					{
						if(i < s1.length() - 1 && s1.charAt(i+1) == '0' && s2.charAt(i+1) == '0')
						{
							res += 2;
							i++;
						}
						else{
							 
						}
					}
					else{
						res += 2;
					}
					 
				}
				System.out.println(res);
			}
			else
			{
				int res = 0;
				for(int i = 0; i < s1.length(); i++)
				{
					if(s1.charAt(i) == '0' && s2.charAt(i) == '0')
						res += 1;
					else if(s1.charAt(i) == '2' && s2.charAt(i) == '2')
					{
						res += 0;
					}
					else{
						res += 1;
					}
				}
				System.out.println(res);
			}
		}
	}
}