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 C {
	
	public static void main(String[] args)throws IOException {
		
		FastScanner scan = new FastScanner();
		BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
		int t = scan.nextInt();
		for(int tt = 0;tt<t;tt++) {
			int n = scan.nextInt();
			String s1 = scan.next();
			String s2 = scan.next();
			int arr1[] = new int[n];
			int arr2[] = new int[n];
			for(int i = 0;i<n;i++) {
				arr1[i] = s1.charAt(i) - '0';
				arr2[i] = s2.charAt(i) - '0';
			}
			int ans = 0;
			for(int i = 0;i<n;i++) {
				if((arr1[i]^arr2[i])==1) {
					ans+=2;
				}
				else if(arr1[i] == 0 && arr2[i] == 0) {
					if((i!=(n-1) && (arr1[i+1] == 1 && arr2[i+1] == 1))) {
						ans+=2;
						arr1[i+1] = -1;
						arr2[i+1] = -1;
						i++;
					}
					else if((i!=0 && (arr1[i-1] == 1 && arr2[i-1] == 1))) {
						ans+=2;
					}
					else
						ans+=1;
				}
			}
			output.write(ans + "\n");
		}
		output.flush();

	}
	
	public static int[] sort(int arr[]) {

		List<Integer> list = new ArrayList<>();
		for(int i:arr)
			list.add(i);
		Collections.sort(list);
		for(int i = 0;i<list.size();i++) {
			arr[i] = list.get(i);
		}
		return arr;

	}
	
	public static int gcd(int a, int b) {
		if(a == 0) return b;
		return gcd(b%a, a);
	}
	
	public static void printArray(int arr[]) {
		for(int i:arr) System.out.print(i+" ");
		System.out.println();
	}
	
	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) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}

}
