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.*;
 
public class C {
public static void main(String[] args) {
		
		FastScanner sc = new FastScanner();
		int t = sc.nextInt();
		for(int tt = 0; tt < t ; tt++) {
			int n = sc.nextInt();
			char a[][] = new char[2][n];
			
			a[0] = sc.next().toCharArray();
			a[1] = sc.next().toCharArray();
			
			int ans = 0;
	outer:		for(int i = 0 ; i < n ; i++) {
				if(i < n-1) {
					if((a[0][i] == '0' && a[1][i] == '0' && a[0][i+1] == '1' && a[1][i+1] == '1') || (a[0][i] == '1' && a[1][i] == '1' && a[0][i+1] == '0' && a[1][i+1] == '0')) {
						ans += 2;
						i++;
						continue outer;
					}
				}
				if(a[0][i] == '0' && a[1][i] == '0')ans += 1;
				if((a[0][i] == '1' && a[1][i] == '0') || (a[0][i] == '0' && a[1][i] == '1')) {
					ans += 2;
				}
			}
			System.out.println(ans);
		}
	}
	
	
	public static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}

  
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }
 
    public static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
    public static long pow(long base, long exp, long MOD) {
        base %= MOD;
        long ret = 1;
        while (exp > 0) {
            if ((exp & 1) == 1) ret = ret * base % MOD;
            base = base * base % MOD;
            exp >>= 1;
        }
        return ret;
    }
 
 
    public static long max(long... ar) {
        long ret = ar[0];
        for (long itr : ar) ret = Math.max(ret, itr);
        return ret;
    }
 
    public static int max(int... ar) {
        int ret = ar[0];
        for (int itr : ar) ret = Math.max(ret, itr);
        return ret;
    }
 
    public static long min(long... ar) {
        long ret = ar[0];
        for (long itr : ar) ret = Math.min(ret, itr);
        return ret;
    }
 
    public static int min(int... ar) {
        int ret = ar[0];
        for (int itr : ar) ret = Math.min(ret, itr);
        return ret;
    }
 
 
    public static long sum(long... ar) {
        long sum = 0;
        for (long itr : ar) sum += itr;
        return sum;
    }
 
    public static long sum(int... ar) {
        long sum = 0;
        for (int itr : ar) sum += itr;
        return sum;
    }
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		
		public String next() {
			while (!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());
		}
		
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
	}
 
}


 
 
 
 
 
 
 
 
