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 q3 {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
        
        int tests = Integer.parseInt(br.readLine());

        for (int test = 1;test <= tests;test++) {
            String[] parts = br.readLine().split(" ");
            int n = Integer.parseInt(parts[0]);

            String[] arr = new String[2];
            for(int i = 0;i < 2;i++){
                arr[i] = br.readLine();
            }

            int ans = 0,i = 0,co = 0,cz = 0;
            while(i < n){
                if(arr[0].charAt(i) == '0') cz++;
                else co++;

                if(arr[1].charAt(i) == '0') cz++;
                else co++;

                if(cz > 0 && co > 0){
                    ans += 2;
                    cz = 0;
                    co = 0;
                }else if(cz > 0){
                    if(i + 1 < n && arr[0].charAt(i + 1) == '1' && arr[1].charAt(i + 1) == '1'){

                    }else{
                        ans++;
                        cz = 0;
                        co = 0;
                    }
                }

                i++;
            }
            out.write(ans + "\n");
        }
        out.flush();
    }
}
