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.
/*package whatever //do not write package name here */

import java.io.*;
import java.util.*;

public class GFG {
	public static void main (String[] args) {
		Scanner scn = new Scanner(System.in);
		int t = scn.nextInt();
		while(t-- > 0){
		  //  int n = scn.nextInt();
		  //  int[] arr = new int[n];
		  //  for(int i=0; i<n; i++){
		  //      arr[i] = scn.nextInt();
		  //  }
		    int n = scn.nextInt();
		    scn.nextLine();
		    
            String[] str = new String[2];
            str[0] = scn.nextLine();
            str[1] = scn.nextLine();
            
            int ans = 0;
            for(int i=0; i<n;){
                
                char ch1 = str[0].charAt(i), ch2 = str[1].charAt(i);
                if(ch1 != ch2){
                    ans += 2;
                    i++;
                }
                else{
                    
                    int tag = 0;
                    while(i < n  &&  str[0].charAt(i) == str[1].charAt(i)){
                        if(str[0].charAt(i) != ch1){
                            ans += 2;
                            tag = 1;
                            i++;
                            break;
                        }
                        if(str[0].charAt(i) == '0'){
                            if(i+1 < n  &&  str[0].charAt(i+1) == str[1].charAt(i+1)){
                                if(str[0].charAt(i+1) == '1'){
                                    ans += 2;
                                    i += 2;
                                    tag = 1;
                                    break;
                                }
                                else{
                                    ans++;
                                    tag = 1;
                                    i++;
                                    break;       
                                }
                            }
                            else{
                                ans++;
                                tag = 1;
                                i++;
                                break;
                            }
                        }
                        i++;
                    }
                    
                    if(tag == 0){
                        if(i == n){
                            if(ch1 == '0'){
                                ans++;
                                i++;
                            }
                        }    
                        else{
                            ans += 2;
                            i++;
                        }
                    }
                }
                // System.out.println(ans + " " + i);
            }
            
            System.out.println(ans);
		}
	}
}