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 A {
    static final Reader s = new Reader();
    static final PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) throws IOException {
        int t  = s.nextInt();
        for(int i=1; i<=t; ++i) {
//            out.print("Case #"+i+": ");
            new Solver();
        }
        out.close();
    }
    static class Solver {
        Solver(){
            int n = s.nextInt();
            int[][] mat = new int[2][n];
            int b=0;
            long cnt=0;
            int last_index=-4;
            int value=4;
            for(int i=0;i<2;i++){
                String temp=s.next();
                char[] g = temp.toCharArray();
                for(int j=0;j<n;j++){
                    mat[b][j] = (int)(g[j]-'0');
                }
                b++;
            }


            for(int i=0;i<n;i++){
                if(mat[1][i]!=mat[0][i]){
                    cnt+=2;
                }
                else{
                    if(mat[0][i]==0){
                        if(last_index==i-1 && value==1){
                            cnt+=2;
                            value=9;
                        }
                        else {
                            cnt++;
                            value=0;
                        }
                        last_index=i;
                    }
                    else {
                        if(last_index==i-1){
                            if(value==0){
                                cnt++;
                                value=9;
                            }
                            else value=1;
                        }
                        else value=1;
                        last_index=i;
                    }
                }
            }
            out.println(cnt);
        }
    }
    static class Reader {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        String next() {
            while(st==null||!st.hasMoreTokens()) {
                try {
                    st=new StringTokenizer(in.readLine());
                } catch(Exception e) {}
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }

}