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.*;

/** 
__                  __
( _)                ( _)
/ / \\              / /\_\_
/ /   \\            / / | \ \
/ /     \\          / /  |\ \ \
/  /   ,  \ ,       / /   /|  \ \
/  /    |\_ /|      / /   / \   \_\
/  /  |\/ _ '_| \   / /   /   \    \\
|  /   |/  0 \0\    / |    |    \    \\
|    |\|      \_\_ /  /    |     \    \\
|  | |/    \.\ o\o)  /      \     |    \\
\    |     /\\`v-v  /        |    |     \\
| \/    /_| \\_|  /         |    | \    \\
| |    /__/_ `-` /   _____  |    |  \    \\
\|    [__]  \_/  |_________  \   |   \    ()
/    [___] (    \         \  |\ |   |   //
|    [___]                  |\| \|   /  |/
/|    [____]                  \  |/\ / / ||
(  \   [____ /     ) _\      \  \    \| | ||
\  \  [_____|    / /     __/    \   / / //
|   \ [_____/   / /        \    |   \/ //
|   /  '----|   /=\____   _/    |   / //
__ /  /        |  /   ___/  _/\    \  | ||
(/-(/-\)       /   \  (/\/\)/  |    /  | /
(/\/\)           /   /   //
_________/   /    /
\____________/    (



*/

public class Main {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		
		while(t-- >0) {
			int n=sc.nextInt();
			String a=sc.next();
			String b=sc.next();
			
			int zo=0;
			int oz=0;
			int ne=0;
			HashSet<Integer> set=new HashSet<>();
			for(int i=0;i<n;i++) {
				char ch1=a.charAt(i);
				char ch2=b.charAt(i);
				
				if(ch1!=ch2 ) {
					ne++;
					set.add(i);
				}
			}
			for(int i=1;i<n;i++) {
				if(a.charAt(i)=='1' && a.charAt(i-1)=='0' && !set.contains(i) && !set.contains(i-1)) {
					set.add(i-1);
					set.add(i);
					zo++;
				}
				else if(a.charAt(i-1)=='1' && a.charAt(i)=='0' && !set.contains(i) && !set.contains(i-1)) {
					set.add(i-1);
					set.add(i);
					oz++;
				}
			}
			int zz=0;
			for(int i=0;i<n;i++) {
				if(a.charAt(i)=='0' && b.charAt(i)=='0' && !set.contains(i)) {
					set.add(i);
					zz++;
				}
			}
			int ans=2*ne;
			ans+=zo*2;
			ans+=oz*2;
			ans+=zz;
			System.out.println(ans);
		}
		
	}
}
