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 tank {

    static final FastScanner fs = new FastScanner();

    //static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        int t = fs.nextInt();
        while(t-- > 0) run_case();

        //out.close();
    }

    static void run_case() {
        int n = fs.nextInt();
        char[] s = fs.next().toCharArray();
        char[] t = fs.next().toCharArray();

        int ans = 0;

        boolean ze = false, on = false;

        for (int i = 0; i < n; i++) {
            if(mex(s[i], t[i]) == 0) {
                if(on) {
                    ans += 2;
                    ze = false;
                    on = false;
                }else {
                    ze = true;
                }
                continue;
            }

            if(mex(s[i], t[i]) == 1) {
                if(on) {
                    ans++;
                }

                if(ze) {
                    ans += 2;
                    on = false;
                    ze = false;
                }else {
                    on = true;
                }
            }

            if(mex(s[i], t[i]) == 2) {
                ans += 2;
                on = false;
                ze = false;
            }
        }

        if(on) {
            ans++;
        }

        System.out.println(ans);

    }

    static int mex(char a, char b) {
        if(a == b) {
            if(a == '1' ) {
                return 0;
            }else {
                return 1;
            }
        }

        return 2;
    }

    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();
        }

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

            return "";
        }

        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());
        }
    }
}