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.*;
import java.util.*;


public class Main {
	public static class FastReader {
		BufferedReader br;
		StringTokenizer root;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (root == null || !root.hasMoreTokens()) {
				try {
					root = new StringTokenizer(br.readLine());
				} catch (Exception r) {
					r.printStackTrace();
				}
			}
			return root.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
	}

		
	public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	static FastReader sc = new FastReader();
	public static int mod = (int)(1e9+7), MAX = 50;
	
	
	public static void main(String args[]) {  
		
		int t = sc.nextInt();
		while(t-->0) {
		    int n = sc.nextInt();
			char[] s = sc.next().toCharArray();
			char[] p = sc.next().toCharArray();
			int cnt = 0, ans = 0;
			for(int i=0;i<n;++i) {
				TreeSet<Integer> set = new TreeSet<>();
				int j = i;
				while(j < n) {
					set.add(s[j]-'0');
					set.add(p[j]-'0');
					if(set.size() == 2) {
						if(cnt > 0) {
							if(s[j] == p[j]) {
								ans+=Math.max(2, cnt);
							}else ans+=cnt+2;
							cnt = 0;
						}else {
							ans+=2;
						}
						i = j;
						break;
						
					}else if(set.size() == 1 && set.first() == 0) ++cnt;
					++j;
				}
				i = Math.max(i, j-1);
				ans+=cnt;
			}
			out.println(ans);
		}
		out.close();
	}
	
	
}