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 template {
    static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;
 
        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public Reader(String file_name) throws IOException
        {
            din = new DataInputStream(
                new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public String readLine() throws IOException
        {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') {
                    if (cnt != 0) {
                        break;
                    }
                    else {
                        continue;
                    }
                }
                buf[cnt++] = (byte)c;
            }
            return new String(buf, 0, cnt);
        }
 
        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ') {
                c = read();
            }
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
 
            if (neg)
                return -ret;
            return ret;
        }
 
        public long nextLong() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }
 
        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
 
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
 
            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }
 
            if (neg)
                return -ret;
            return ret;
        }
 
        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0,
                                 BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }
 
        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }
 
        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }
 
    public static void main(String[] args)
        throws IOException
    {
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        while (t-->0) {
        int flag = 0;
        long n = s.nextInt();
        String s1 = s.next();
        String s2 = s.next();
        int ans =0;
        for(int i=0;i<s1.length();i++)
        {
             if(s1.charAt(i)=='1' && s2.charAt(i)=='1')
        {
            while(i<s1.length() && (s1.charAt(i)=='1' && s2.charAt(i)=='1'))
            {
                i++;
            }
            if(!(i==s1.length()))
            {
                ans+=2;
            }
        }
        else
        {
            if(s1.charAt(i)=='0' && s2.charAt(i)=='0')
            {
                if(i+1<s1.length() && s1.charAt(i+1)=='1' && s2.charAt(i+1)=='1')
                {
                    ans+=2;
                   i++;
                }
                else
                    ans++;
            }
            else
            {
                ans=ans+2;
            }
        }
        }
      System.out.println(ans);
        
      
    }
    }}