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.util.*;
import java.io.*;

public class test {

    static long mod = 1000000007;
    static long inf = Long.MAX_VALUE;
    static final int maxSizeSeive = 1000001;
    static Vector<Boolean>isprime = new Vector<>(maxSizeSeive);
    static Vector<Integer>prime = new Vector<>();
    static Vector<Integer>SPF = new Vector<>(maxSizeSeive);

    public static void solve(int n, char[] s1, char[] s2) {
        boolean visited[] = new boolean[n];
        int[] group2 = new int[n];
        int numberOfGrp2 = 0;
        int[] group1 = new int[n];
        int[] group0 = new int[n];
        for(int i = 0; i < n; i++) {
            if(s1[i] == '1' && s2[i] == '0') {
                group2[i] = 1;
                numberOfGrp2 += 1;
                visited[i] = true;
            }
            else if(s2[i] == '1' && s1[i] == '0') {
                group2[i] = 1;
                numberOfGrp2 += 1;
                visited[i] = true;
            }
            else if(s2[i] == '1' && s1[i] == '1') {
                group1[i] = 1;
            }
            else if(s2[i] == '0' && s1[i] == '0') {
                group0[i] = 1;
            }
        }
        //System.out.println("test");
        if(numberOfGrp2 == n) {
            System.out.println((n*2));
            return;
        }
        //System.out.println("number of grp 2 : " + numberOfGrp2);

        /*for(int i  : group0) {
            System.out.print(i);
        }
        System.out.println();
        for(int i : group1) {
            System.out.print(i);
        }
        System.out.println();*/
        /*System.out.println("grping");

        int[] grouping = new int[n];
        for(int i = 0; i < n; i++) {
            grouping[i] = group0[i] + group1[i];
            System.out.print(grouping[i]);
        }
        System.out.println();*/
        
        for(int i = 1; i < n; i++) {
            if(!visited[i] && !visited[i - 1]) {
                if(group0[i] == 1 && group1[i - 1] == 1) {
                    visited[i] = true;
                    visited[i - 1] = true;
                    numberOfGrp2 += 1;
                }
                else  if(group1[i] == 1 && group0[i - 1] == 1) {
                    visited[i] = true;
                    visited[i - 1] = true;
                    numberOfGrp2 += 1;
                }
            }
        }

        //System.out.println("number of grp 2 : " + numberOfGrp2);
        long sum = numberOfGrp2 * 2;
        for(int i = 0; i < n; i++) {
            if(!visited[i]) {
                if(group0[i] == 1) {
                    sum += 1;
                }
            }
        }

        System.out.println(sum);
    }
    
/*
input:
4
7
0101000
1101100
5
01100
10101
2
01
01
6
000000
111111

8
10011100
01011100

output:
8
8
2
12

*/

    public static void main(String[] args) {
        FastReader sc = new FastReader();
        int t = sc.nextInt();
        while(t-- > 0) {
            int n = sc.nextInt();
            char[] s1 = sc.nextLine().toCharArray();
            char[] s2 = sc.nextLine().toCharArray();
            // solve
            solve(n, s1, s2);
        }
    }

    public static ArrayList<Integer> seivePrime(int n) {
        ArrayList<Integer> primes = new ArrayList<>();
        for (int i = 0; i < maxSizeSeive; i++){
            isprime.add(true);
            SPF.add(2);
        }

        manipulated_seive(n);

        for (int i=0; i<prime.size() && prime.get(i) <= n; i++) {
            primes.add(prime.get(i));
        }
        return primes;
    }

    public static void manipulated_seive(int N) {
        isprime.set(0, false);
        isprime.set(1, false);
         
        for (int i=2; i<N ; i++) {
            if (isprime.get(i)) {
                prime.add(i);
                SPF.set(i,i);
            }

            for (int j=0;j < prime.size() && i*prime.get(j) < N && prime.get(j) <= SPF.get(i);j++) {
                isprime.set(i*prime.get(j),false);
                SPF.set(i*prime.get(j),prime.get(j));
            }
        }
    }    
     
	public static void sort(long[] a) {
		ArrayList<Long> l=new ArrayList<>();
		for (long i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}

    public static int maxIndex(long[] arr) {
        long max = Long.MIN_VALUE;
        int index = 0;
        int n = arr.length;
        for(int i = 0; i < n; i++) {
            if(arr[i] > max) {
                index = i;
                max = arr[i];
            }
        }
        return index;
    }

    public static int minIndex(long[] arr) {
        long min = Long.MAX_VALUE;
        int index = 0;
        int n = arr.length;
        for(int i = 0; i < n; i++) {
            if(arr[i] < min) {
                index = i;
                min = arr[i];
            }
        }
        return index;
    }

    public static void debugArray(long[] arr) {
        for(int i = 0; i < arr.length; i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();
    } 

    public static void debugArrayList(ArrayList<Long> arr) {
        for(Long i : arr) {
            System.out.print(i + " ");
        }
        System.out.println();
    }

    public static long gcd(long a, long b) {
        if(b == 0) {
            return a;
        }
        else {
            return gcd(b, a % b);
        }
    }
 
    public static long LCM(long a, long b) {
        return Math.abs(a*b) / gcd(a, b);
    }

    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();
        }

        String nextLine() {
            String str="";
            try {
                str=br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

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

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

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

        long[] readArray(int n) {
            long[] a = new long[n];
            for(int i = 0; i < n ; i++) {
                a[i] = nextLong();
            }
            return a;
        }
    }
}