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.
//package codeforce800;

import java.util.Scanner;
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 * ��Ϊ0����00������1������������2���ϲ�
 * ��Ϊ2����11������0������������0���ϲ�
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 * */
public class Max_Mex {
	static int mex(int i,int j) {
		if(i+j==2) return 0;
		else return i+j+1;
	}
	
	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		int times=in.nextInt();
		for (int i=1;i<=times;i++) {
			int length=in.nextInt();
			in.nextLine();
			int sum=0;
			char[] up=in.nextLine().toCharArray();
			char[] down=in.nextLine().toCharArray();
			for(int j=0;j<=length-1;) {
				int mex=mex((int)up[j]-48,(int)down[j]-48);
				if(j==length-1) {
					sum+=mex;
					break;
				}
				if(mex==2) {
					sum+=2;
					j++;
				}else if(mex==1){
					if((int)up[j+1]-48+(int)down[j+1]-48==2) {
						sum+=2;
						j+=2;
					}else {
						sum++;
						j++;
					}
				}else {
					if((int)up[j+1]-48+(int)down[j+1]-48==0) {
						sum+=2;
						j+=2;
					}else j++;
				}
			}
			System.out.println(sum);
		}
	}
}
