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.IOException;
import java.io.InputStreamReader;
import java.util.*;

	public class practice{
		public static void main(String[] args){
			FastScanner r = new FastScanner();					
			int t = r.i() ,n;
			char[][] mat;
			int max=0;
			boolean[] vis;
			StringBuffer ans = new StringBuffer("");
			while(t-->0) {
				n = r.i();
				mat  = new char[2][n];
				mat[0] = r.n().toCharArray();
				mat[1] = r.n().toCharArray();
				max=0;
				vis = new boolean[n];
				for(int i=0;i<n;i++) {
					if(mat[0][i] != mat[1][i]) {
						max+=2;
					}
					else if(mat[0][i]=='0') {
						
						
						if( i-1>=0&&!vis[i-1] && mat[0][i-1] == mat[1][i-1] && mat[1][i-1]=='1') {
							vis[i]=true;
						max+=2;}
						else
								max++;
					}
					
					else {
						if( i-1>=0&&!vis[i-1] && mat[0][i-1] == mat[1][i-1] && mat[1][i-1]=='0')
							max+=1;
					}
//					p(max + " c");
					
				}
				ans.append(max + "\n");
			}				
			p(ans);
		}

		
		static <T> void p(T s) {
			System.out.println(s);
		}
		
		static void p() {
			System.out.println();
		}		
		static <T> void pn(T s) {
			System.out.print(s);
		}

	}
		

	
	class FastScanner {
	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    StringTokenizer st=new StringTokenizer("");
	    String n() {
	        while (!st.hasMoreTokens())
	            try {
	                st=new StringTokenizer(br.readLine());
	            } catch (IOException e) {
	                e.printStackTrace();
	            }
	        return st.nextToken();
	    }
	    int[] ia(int n) {
	    	int[] arr = new int[n];
	    	for(int i=0;i<n;i++)
	    		arr[i] = i();
	    	return arr;
	    }
	    String[] sa(int n) {
	    	String[] arr = new String[n];
	    	for(int i=0;i<n;i++)
	    		arr[i] = n();
	    	return arr;
	    }
	    long[] la(int n) {
	    	long[] arr = new long[n];
	    	for(int i=0;i<n;i++)
	    		arr[i] = l();
	    	return arr;
	    }
	    String nl()
	    {
	        String str = "";
	        try
	        {
	            str = br.readLine();
	        }
	        catch (IOException e)
	        {
	            e.printStackTrace();
	        }
	        return str;
	    }

	    int i() {
	        return Integer.parseInt(n());
	    }
	    long l() {
	        return Long.parseLong(n());
	    }
	}
