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.InputStreamReader;
import java.util.*;

public class MEX2 {
	public static void main(String[] args) throws Exception{
		InputStreamReader inp = new InputStreamReader(System.in);
		BufferedReader buff = new BufferedReader(inp);
		int cases = Integer.parseInt(buff.readLine());
		for(int i = 0; i< cases; i++) {
			int sum = 0;
			int length = Integer.parseInt(buff.readLine());
			List<String> strings = new ArrayList<String>();
			String bin = buff.readLine();
			String bin2 = buff.readLine();
			int start = 0, end = 0;
			for(int j = 0; j<bin.length(); j++) {
				if(j == bin.length() -1) {
					if(bin.charAt(j) != bin2.charAt(j)) {
						end = j;
						if(end!=start) {
							strings.add(bin.substring(start, end));
						}
						start = j+1;
						sum +=2;
					}else {
						strings.add(bin.substring(start));
					}
				}else {
					if(bin.charAt(j) != bin2.charAt(j)) {
						end = j;
						if(end!=start) {
							strings.add(bin.substring(start, end));
						}
						start = j+1;
						sum +=2;
					}
				}
				
				
			}
			for(String s: strings) {
				sum+= findMax(s);
			}

			System.out.println(sum);
		}
	}
	
	public static int findMax(String s) {
		int sum = 0; 
		for(int i = 0; i<s.length(); i++) {
			if(i == s.length()-1) {
				if(Integer.parseInt(s.substring(i, i+1)) == 1){
					continue;
				}else {
					sum ++;
				}
			}else {
				if(Integer.parseInt(s.substring(i, i+1)) == 0 && Integer.parseInt(s.substring(i, i+1)) == Integer.parseInt(s.substring(i+1, i+2))) {
					sum ++;
				}else if(Integer.parseInt(s.substring(i, i+1)) == 1 && Integer.parseInt(s.substring(i, i+1)) == Integer.parseInt(s.substring(i+1, i+2))) {
					continue;
				}else if(Integer.parseInt(s.substring(i, i+1)) == 0 && Integer.parseInt(s.substring(i, i+1)) != Integer.parseInt(s.substring(i+1, i+2))) {
					sum += 2; 
					i++;
				}else if(Integer.parseInt(s.substring(i, i+1)) == 1 && Integer.parseInt(s.substring(i, i+1)) != Integer.parseInt(s.substring(i+1, i+2))) {
					sum += 2; 
					i++;
				}
			}
			
		}
		return sum;
	}
}