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.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.*;
import java.io.*;
public class Main {
    public static void main(String args[])throws Exception{
        Input sc=new Input();
        StringBuilder sb=new StringBuilder();
        int t=sc.readInt();
        for(int f=0;f<t;f++){
            int n=sc.readInt();
            String str[]=new String[2];
            str[0]=sc.readString();
            str[1]=sc.readString();
            int arr[]=new int[n];
            for(int i=0;i<n;i++){
                int v1=str[0].charAt(i)-'0';
                int v2=str[1].charAt(i)-'0';
                arr[i]=v1+v2;
            }
            int sum=0;
//            for(int i=0;i<n;i++){
//                System.out.print(arr[i]+" ");
//            }
//            System.out.println();
            for(int i=0;i<n;i++){
                if(arr[i]==0){
                    if(i>=1 && arr[i-1]==2)
                        arr[i-1]=-1;
                    else if(i<n-1 && arr[i+1]==2)
                        arr[i+1]=-1;
                }
            }
            for(int i=0;i<n;i++){
                if(arr[i]==0)
                    arr[i]=1;
                else if(arr[i]==1)
                    arr[i]=2;
                else if(arr[i]==-1)
                    arr[i]=1;
                else if(arr[i]==2)
                    arr[i]=0;
                sum+=Math.abs(arr[i]);
            }
            if(arr[n-1]==-1)
                sum++;
            sb.append(sum+"\n");
        }
        System.out.print(sb);
    }
}
class Input{
    BufferedReader br;
    StringTokenizer st;
    Input(){
        br=new BufferedReader(new InputStreamReader(System.in));
        st=new StringTokenizer("");
    }
    public int[] readArray() throws Exception{
        st=new StringTokenizer(br.readLine());
        int a[]=new int[st.countTokens()];
        for(int i=0;i<a.length;i++){
            a[i]=Integer.parseInt(st.nextToken());
        }
        return a;
    }
    public long[] readArrayLong() throws Exception{
        st=new StringTokenizer(br.readLine());
        long a[]=new long[st.countTokens()];
        for(int i=0;i<a.length;i++){
            a[i]=Long.parseLong(st.nextToken());
        }
        return a;
    }
    public int readInt() throws Exception{
        st=new StringTokenizer(br.readLine());
        return Integer.parseInt(st.nextToken());
    }
    public long readLong() throws Exception{
        st=new StringTokenizer(br.readLine());
        return Long.parseLong(st.nextToken());
    }
    public String readString() throws Exception{
        return br.readLine();
    }
    public int[][] read2dArray(int n,int m)throws Exception{
        int a[][]=new int[n][m];
        for(int i=0;i<n;i++){
            st=new StringTokenizer(br.readLine());
            for(int j=0;j<m;j++){
                a[i][j]=Integer.parseInt(st.nextToken());
            }
        }
        return a;
    }
}


