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.io.*;
import java.util.*;
public class dhet {

	public static void main(String[] args) throws IOException {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			String s1=sc.next();
			String s2=sc.next();
			long sum=0;
			for(int i=0;i<n;i++) {
				if((s1.charAt(i)=='0' && s2.charAt(i)=='1') 
						|| (s2.charAt(i)=='0' && s1.charAt(i)=='1')) {
					sum+=2;
					//System.out.println(i);
				}
				else if(s1.charAt(i)=='1' && s2.charAt(i)=='1') {
					//System.out.println(i);
					for(int j=i+1;j<n;j++) {
						if(s1.charAt(j)=='0' && s2.charAt(j)=='0') {
							sum+=2;
							i++;
							break;
						}
						else if((s1.charAt(j)=='0' && s2.charAt(j)=='1') 
								|| (s2.charAt(j)=='0' && s1.charAt(j)=='1')) {
							sum+=2;
							i++;
							break;
						}
						else {
							i++;
						}
					}
				}
				else {
					//System.out.println(i);
					if(i+1==n) {
						sum++;
					}
					else if(i+1<n && (s1.charAt(i+1)=='0' && s2.charAt(i+1)=='0')) {
						sum++;
					}
					else if(i+1<n && (s1.charAt(i+1)=='0' && s2.charAt(i+1)=='1') 
							|| (s2.charAt(i+1)=='0' && s1.charAt(i+1)=='1')) {
						sum++;
					}
					else if(i+1<n && (s1.charAt(i+1)=='1' && s2.charAt(i+1)=='1')) {
						sum+=2;
						i++;
					}
					
					//System.out.println(sum+"sum");
				}
				
			}
			System.out.println(sum);
		}
		//System.out.println();
		//System.out.println();
	}
	
}
