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.Collections;
import java.util.StringTokenizer;

public class C implements Runnable {

	public static void main(String[] args) throws Exception {

		new Thread(null, new C(), "SwapnilGanguly", 1 << 20).start();
	}

	public void run() {

		SwapScanner sc = new SwapScanner();
		PrintWriter pr = new PrintWriter(System.out);

		int T = sc.nextInt();

		while (T-- > 0) {

			int n = sc.nextInt();
			char[][] s = new char[2][n];

			s[0] = sc.next().toCharArray();
			s[1] = sc.next().toCharArray();

			int ans = 0;

			for (int i = 0; i < n; i++) {

				if (s[0][i] != s[1][i])
					ans += 2;

				else {

					if (i < n - 1 && s[0][i + 1] == s[1][i + 1]) {

						if (s[0][i] == '0' || s[0][i + 1] == '0') {

							i++;
							ans += 2;
						}
					}

					else {

						if (s[0][i] == '0')
							ans++;

					}
				}
			}

			pr.println(ans);
		}

		pr.close();
	}

	public static void sort(int[] a) {

		ArrayList<Integer> l = new ArrayList<>();

		for (int i : a)
			l.add(i);

		Collections.sort(l);

		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}

	public static class SwapScanner {

		BufferedReader br;
		StringTokenizer st;

		public SwapScanner() {

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

		String next() {

			while (st == null || !st.hasMoreElements()) {

				try {

					st = new StringTokenizer(br.readLine());
				}

				catch (IOException e) {

					e.printStackTrace();
				}
			}

			return st.nextToken();
		}

		int nextInt() {

			return Integer.parseInt(next());
		}

		long nextLong() {

			return Long.parseLong(next());
		}

		double nextDouble() {

			return Double.parseDouble(next());
		}

		String nextLine() {

			String str = "";

			try {

				str = br.readLine();
			}

			catch (IOException e) {

				e.printStackTrace();
			}

			return str;
		}

		int[] readArray(int n) {

			int[] ar = new int[n];

			for (int i = 0; i < n; i++)
				ar[i] = nextInt();

			return ar;
		}
	}
}