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 MaxMexCut {

    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int t = in.nextInt();
        for (int i = 1; i <= t; i++)  {
        	int n = in.nextInt();
        	String s1 = in.next();
        	String s2 = in.next();
        	
        	char[] s1Arr = s1.toCharArray();
        	char[] s2Arr = s2.toCharArray();
        	
        	int result = 0;
        	for (int j = 0; j < n; j++)  {
        		if (s1Arr[j] != s2Arr[j])  {
        			result += 2;
        		}
        		else  {
        			if ((s1Arr[j] == '0') && (s2Arr[j] == '0'))  {
        				result += 1;
        				if (j > 0)  {
            				if ((s1Arr[j - 1] == '1') && (s2Arr[j - 1] == '1'))  {
            					result += 1;
            				}
            				else  {
                				if (j < n - 1)  {
                    				if ((s1Arr[j + 1] == '1') && (s2Arr[j + 1] == '1'))  {
                    					result += 1;
                    				}
                				}
            				}
        				}
        				else  {
            				if (j < n - 1)  {
                				if ((s1Arr[j + 1] == '1') && (s2Arr[j + 1] == '1'))  {
                					result += 1;
                				}
            				}
        				}
        			}
        		}
        	}
        	out.println(result);
        }
        out.close();
    }
    
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try { 
                    st = new StringTokenizer(br.readLine());                               
                } catch (IOException e) {
                    // noop
                }
            return st.nextToken();
        }
        
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble()  {
            return Double.parseDouble(next());
        }
    }

}