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.
    /*
     * To change this license header, choose License Headers in Project Properties.
     * To change this template file, choose Tools | Templates
     * and open the template in the editor.
     */
    import java.util.*;
    import java.util.Arrays;


    /**
     *
     * @author DELL
     */
      public class solution{

               static long power(long n,long ans){
                   if(n==0)return ans;
                   return ans=power(n-1,ans)*2;
               }
               static boolean prime(int n){
                   for(int i=2;i*i<=n;i++){
                       if(n%i==0)return false;
                   }
                   return true;
               }
               static long power(long x, int y, int p)
                {
                  long res = 1; // Initialize result

                  x = x % p; // Update x if it is more than or
                  // equal to p

                  if (x == 0)
                    return 0; // In case x is divisible by p;

                  while (y > 0)
                  {

                    // If y is odd, multiply x with result
                    if ((y & 1) != 0)
                      res = (res * x) % p;

                    // y must be even now
                    y = y >> 1; // y = y/2
                    x = (x * x) % p;
                  }
                  return res;
                }
               static int gcd(int a,int b)
                {
                    if(b%a==0)
                    return a;
                    return gcd(b%a,a);
                }

        public static void main(String[] args) {
             Scanner sc=new Scanner(System.in);
             StringBuilder sb=new StringBuilder();
             int t=sc.nextInt();
             while(t-->0){
                 int n=sc.nextInt();
                 String a=sc.next();
                 String b=sc.next();
                 int ans=0;
                 int arr[]=new int[n];
                 for(int i=0;i<a.length();i++){
                     if(a.charAt(i)!=b.charAt(i))ans+=2;
                     else if(a.charAt(i)=='1'){
                         if(i-1>=0 && arr[i-1]==0 && a.charAt(i-1)=='0' && b.charAt(i-1)=='0'){
                             ans++;
                             arr[i-1]=1;
                         }
                         else if(i+1<a.length() && a.charAt(i+1)=='0' && b.charAt(i+1)=='0' && arr[i+1]==0){
                             ans++;
                             arr[i+1]=1;
                         }
                     }
                     else ans++;
                 }
                 System.out.println(ans);
             }
        }

    }
