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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.*;
public class c731 {

	public static void main(String[] args) throws IOException {
//try {
FastReader sc = new FastReader();
BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
int t = sc.nextInt();

for(int o = 0 ; o<t;o++) {
int n = sc.nextInt();
String s1 = sc.next();
String s2 = sc.next();
long ans = 0;
for(int i =0; i<n;i++) {
	if(s1.charAt(i) == '0' && s2.charAt(i) == '1') {
		ans += 2;
		
	}else if(s1.charAt(i) == '1' && s2.charAt(i) == '0') {
		ans+= 2;
	}else if(s1.charAt(i) == '0' && s2.charAt(i) == '0') {
		if(i == n-1) {
			ans ++;
		}else {
			if(s1.charAt(i+1) == '0' && s2.charAt(i+1) == '1') {
				ans += 3;
				i++;
			}else if(s1.charAt(i+1) == '1' && s2.charAt(i+1) == '0') {
				ans+= 3;
			i++;
			}else if(s1.charAt(i+1) == '1' && s2.charAt(i+1) == '1') {
				ans+=2;	
		i++;
			}else {
				ans++;
			}
		}
		
	}else {
//		int j = i+1;
//		while(j<n) {
//			if(s1.charAt(j) == '0' && s2.charAt(j) == '1') {
//				ans += 2;
//				break;
//			}else if(s1.charAt(j) == '1' && s2.charAt(j) == '0') {
//				ans += 2;
//				break;
//			}else if(s1.charAt(j) == '0' && s2.charAt(j) == '0') {
//				ans+=2;
//				break;
//			}
//		j++;
//		}
//	i = j;
	if(i == n-1) {
		continue;
	}else {
	if(s1.charAt(i+1) == '1' && s2.charAt(i+1) == '1') {
		continue;
	}else {
		ans+=2;
	i++;
	}
	}
	
	}
}
System.out.println(ans);
}


}

	
	
	public static long add(long a , long b , int m) {
		return(a%m + b%m + m)%m;
	}
	public static long mul(long a , long b , int m) {
		return (a* b) %m;
	}
	
	public static long power(long x, int y, int p)
	    {
	 
	        long res = 1;
	 
	        x = x % p;
	 
	        while (y > 0) {
	 
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	 
	        return res;
	    }
	 
	    public static long modInverse(long n, int p)
	    {
	        return power(n, p - 2, p);
	    }
	 
	   public static long ncr(int n, int r,
	                             int p,long[] fac)
	    {
	 
	          if (n<r)
	              return 0;
	        if (r == 0)
	            return 1;
	 
	 
	        return (fac[n] * modInverse(fac[r], p)
	                % p * modInverse(fac[n - r], p)
	                % p)
	            % p;
	    }
	 public static  void shuffleArray(long[] arr){
	        int n = arr.length;
	        Random rnd = new Random();
	        for(int i=0; i<n; ++i){
	            long tmp = arr[i];
	            int randomPos = i + rnd.nextInt(n-i);
	            arr[i] = arr[randomPos];
	            arr[randomPos] = tmp;
	        }   
	}

}


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;
    }
}