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 A17 
{
	public static void process()throws IOException
	{
		
		int n=I();
	 String s=S();
	 
      char ch[]=s.toCharArray();
      String s1=S();
      char ch1[]=s1.toCharArray();
      int max=0;
     for(int i=0;i<n;i++)
     {
    	 if(ch[i]=='0' && ch1[i]=='0')
    	 {
    		 
    		 int k=i;
    		 while(k<n && ch[k]=='0' && ch1[k]=='0')
    		 {
    			 k++;
    		 }
//    		 pn(i+" "+k);
   		    if(k==n)
   		    {
   		    	max+=(k-i);
   		    	i=k;
   		    }
   		    else
   		    {
   		    	if(ch[k]=='1'&&ch1[k]=='1')
   		    	{
   		    		max+=k-i-1+2;
   		    	}
   		    	else
   		    	{
   		    		max+=(k-i)+2;
   		    	}
   		    	i=k;
   		    }
    	 }
    	 else
    	  if(ch[i]=='0' || ch1[i]=='0')
    	 {
    		
    		 max+=2;
//    		  pn(i+" 2 "+" "+max);
    	 }
    	  else
    	 if(ch[i]=='1' &&  ch[i]=='1')
    	 {
    		 int k=i;
    		 while(k<n && ch[k]=='1' && ch1[k]=='1')
    		 {
    			 k++;
    		 }
    		 if(k>=n)
    		 {
    			 max+=0;
    		 }
    		 else
    		 {
    			 max+=2;
    		 }
//    		  pn(i+" 3 "+k+" "+max);
    		 i=k;
    	 }
     }
     pn(max);
     
	}
	static Scanner sc = new Scanner(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	static void pn(Object o){out.println(o);out.flush();}
	static void p(Object o){out.print(o);out.flush();}
	static void pni(Object o){out.println(o);System.out.flush();}
	static int I() throws IOException{return sc.nextInt();}
	static long L() throws IOException{return sc.nextLong();}
	static double D() throws IOException{return sc.nextDouble();}
	static String S() throws IOException{return sc.next();}
	static char C() throws IOException{return sc.next().charAt(0);}
	static int[] Ai(int n) throws IOException{int[] arr = new int[n];for (int i = 0; i < n; i++)arr[i] = I();return arr;}
	static String[] As(int n) throws IOException{String s[] = new String[n];for (int i = 0; i < n; i++)s[i] = S();return s;}
	static long[] Al(int n) throws IOException {long[] arr = new long[n];for (int i = 0; i < n; i++)arr[i] = L();return arr;}
	static void dyn(int dp[][],int n,int m,int z)throws IOException {for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ dp[i][j]=z;}} }
	
//	*--------------------------------------------------------------------------------------------------------------------------------*//
	static class AnotherReader {BufferedReader br;StringTokenizer st;AnotherReader() throws FileNotFoundException {br = new BufferedReader(new InputStreamReader(System.in));}
	AnotherReader(int a) throws FileNotFoundException {br = new BufferedReader(new FileReader("input.txt"));}
	String next() throws IOException{while (st == null || !st.hasMoreElements()) {try {st = new StringTokenizer(br.readLine());} catch (IOException e) {e.printStackTrace();}}return st.nextToken();}
	int nextInt() throws IOException {return Integer.parseInt(next());}
	long nextLong() throws IOException {return Long.parseLong(next());}
	double nextDouble() throws IOException {return Double.parseDouble(next());}
	String nextLine() throws IOException {String str = "";try {str = br.readLine();} catch (IOException e) {e.printStackTrace();}return str;}}
	public static void main(String[] args)throws IOException{try{boolean oj=true;if(oj==true)
	{AnotherReader sk=new AnotherReader();PrintWriter out=new PrintWriter(System.out);}
	else
	{AnotherReader sk=new AnotherReader(100);out=new PrintWriter("output.txt");}
	long T=L();while(T-->0)
	{process();}out.flush();out.close();}catch(Exception e){return;}}}
//*-----------------------------------------------------------------------------------------------------------------------------------*//
	
