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.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.PrintStream;
import java.util.Arrays;
import java.util.HashSet;
public class Sol3 
{ 
	public static  FastReader fs;
	public static void main(String[] args) 
	{ 
		try {
			System.setIn(new FileInputStream("input.txt"));
			System.setOut(new PrintStream(new FileOutputStream("output.txt")));
		} catch (Exception e) {}
		fs = new FastReader();
		int test = fs.nextInt();
		while(test-- > 0){
			solve();
		}
	}

	static class FastReader 
	{ 
		BufferedReader br; 
		StringTokenizer st; 

		public FastReader() 
		{ 
			br = new BufferedReader(new
					InputStreamReader(System.in)); 
		} 

		String next() 
		{ 
			while (st == null || !st.hasMoreElements()) 
			{ 
				try
				{ 
					st = new StringTokenizer(br.readLine()); 
				} 
				catch (IOException e) 
				{ 
					e.printStackTrace(); 
				} 
			} 
			return st.nextToken(); 
		} 

		int nextInt() 
		{ 
			return Integer.parseInt(next()); 
		} 

		long nextLong() 
		{ 
			return Long.parseLong(next()); 
		} 

		double nextDouble() 
		{ 
			return Double.parseDouble(next()); 
		} 

		String nextLine() 
		{ 
			String str = ""; 
			try
			{ 
				str = br.readLine(); 
			} 
			catch (IOException e) 
			{ 
				e.printStackTrace(); 
			} 
			return str; 
		} 
	} 
	// use fs as scanner
	// ctrl + shift + b to compile and run 
	public static void solve(){
		// start code here
		int len = fs.nextInt();
		String[] arr = new String[2];

		arr[0] = fs.nextLine();
		arr[1] = fs.nextLine();
		int sum = 0;
		for(int i=0; i<len; i++){
			char next1 = i == len - 1 ? '5' : arr[0].charAt(i+1);
			char next2 = i == len - 1 ? '5' : arr[1].charAt(i+1);
			if(arr[0].charAt(i) == '0' && arr[1].charAt(i) == '1') {
				sum += 2;
			}
			else if(arr[0].charAt(i) == '1' && arr[1].charAt(i) == '0'){
				sum += 2;
			}
			else if(arr[0].charAt(i) == '0' && arr[1].charAt(i) == '0'){
				if( next1 == '0' && next2 == '1' ){
					sum += 1;
				}
				else if(next1 == '1' && next2 == '0' ){
					sum += 1;
				}
				else if(next1 == '1' && next2 == '1'){
					sum += 2;
					i++;
				}
				else{
					sum += 1;
				}
			}
			else{
				if( next1 == '0' && next2 == '1' ){
					sum += 0;
					// i++;
				}
				else if(next1 == '1' && next2 == '0' ){
					sum += 0;
					// i++;
				}
				else if(next1 == '1' && next2 == '1'){
					sum += 0;
				}
				else{
					sum += 2;
					i++;
				}
			}
			// System.out.print(sum+" ");
		}
		System.out.println(sum);	

	}
}