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.Scanner;

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();
             sc.nextLine();
             String s1 = sc.nextLine();
             String s2 = sc.nextLine();
             char[][] table = new char[2][n];
             for(int i=0; i<n; i++)
             {
              table[0][i] = s1.charAt(i);
              table[1][i] = s2.charAt(i);
             }
             int sum = 0;
             if(n<2)
             {
                 if(table[0][0]=='1' && table[1][0]=='1')
                 {
                     System.out.println(0);
                 }
                 else if(table[0][0]=='0' && table[1][0]=='0')
                 {
                     System.out.println(1);
                 }
                 else
                 {
                     System.out.println(2);
                 }
             }
             else {
                 for (int i = 0; i <=n-1; ) {
                     if (table[0][i] == '0' && table[1][i] == '0') {
                         if(i<n-1) {
                             if ((table[0][i + 1] == '1' && table[1][i + 1] == '1')) {
                                 sum = sum + 2;
                                 i = i + 2;
                             } else if ((table[0][i + 1] == '0' && table[1][i + 1] == '0')) {
                                 sum = sum + 1;
                                 i = i + 1;
                             } else {
                                 sum = sum + 1 + 2;
                                 i = i + 2;
                             }
                         }
                         else
                         {
                             sum = sum+1;
                             i++;
                         }
                     } else if (table[0][i] == '1' && table[1][i] == '1') {
                         if(i<n-1) {
                             if ((table[0][i + 1] == '0' && table[1][i + 1] == '0')) {
                                 sum = sum + 2;
                                 i = i + 2;
                             } else {
                                 i++;
                             }
                         }
                         else{
                             i++;
                         }
                     } else {
                         sum = sum + 2;
                         i++;
                     }

                 }
                 System.out.println(sum);
//                 sc.nextLine();
             }
         }
    }
}