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.io.PrintWriter;
//import java.util.ArrayList;
import java.util.StringTokenizer;

public class radix {

	void solve() throws IOException {

		int n = in.nextInt();
		String S1;
		String S2;

		for (int g = 0; g < n; g++) {
			in.nextInt();
			S1 = in.next();
			S2 = in.next();

			int count = 0;
			char start = 'S';
			for (int i = 0; i < S1.length(); i++) {
				if (S1.charAt(i) != S2.charAt(i)) {
					count += 2;
					if (start == '0') {
						count++;
					}
					start = 'S';
				} else {
					if (start == '0' && S1.charAt(i) == '0') {
						count += 1;
					}
					if (start != S1.charAt(i)) {
						if (start != 'S') {
							count += 2;
							start = 'S';
						} else {
							start = S1.charAt(i);
						}
					}
				}
			}
			if (start == '0') {
				count++;
			}
			out.println(count);
		}
	}

	FastScanner in;
	static PrintWriter out;

	public void run() {
		try {
			in = new FastScanner();
			out = new PrintWriter(System.out);

			solve();

		} catch (Throwable e) {
			e.printStackTrace();
		} finally {
			out.close();
		}
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		boolean eof;

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

		boolean hasNext() {
			if (st == null || !st.hasMoreTokens()) {
				return false;
			}
			return true;
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					eof = true;
					return null;
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			String s = next();
			return eof ? null : Integer.parseInt(s);
		}
	}

	public static void main(String[] arg) {
		new radix().run();
	}

}