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.util.*;
import java.io.*;
public class q {

	static class fast {
		BufferedReader br;
		StringTokenizer st;
		public fast() {
			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;
		}
	}

	public static void main(String[] args)throws IOException {
		if (System.getProperty("ONLINE_JUDGE") == null) {
			try {
				System.setIn(new FileInputStream(new File("i.txt")));
				System.setOut(new PrintStream(new File("o.txt")));
			} catch (Exception e) {}
		}
		Scanner f = new Scanner(System.in);
		int t = f.nextInt();
		StringBuilder sb = new StringBuilder();
		while (t-- > 0) {
			int n = f.nextInt();
			String s1 = f.next();
			String s2 = f.next();
			int[]arr = new int[n];
			for (int i = 0; i < n; i++) {
				if (s1.charAt(i) == '0' && s2.charAt(i) == '0') {
					arr[i] = 1;
					continue;
				}
				if (s1.charAt(i) == '1' && s2.charAt(i) == '0') {
					arr[i] = 2;
					continue;
				}
				if (s1.charAt(i) == '0' && s2.charAt(i) == '1') {
					arr[i] = 2;
					continue;
				}
				if (s1.charAt(i) == '1' && s2.charAt(i) == '1') {
					arr[i] = 0;
					continue;
				}
			}
			long ans = 0;
			for (int i = 0; i < n; i++) {
				if (arr[i] == 2) {
					ans += 2;
					continue;
				}
				if (arr[i] == 1 && i < n - 1 && arr[i + 1] == 0) {
					ans += 2;
					i++;
					continue;
				} else
					if (arr[i] == 1 && i < n - 1 && arr[i + 1] != 0) {
						ans += 1;
						continue;
					}
				if (arr[i] == 0 && i < n - 1 && arr[i + 1] == 1) {
					ans += 2;
					i++;
					continue;
				} else
					if (arr[i] == 0 && i < n - 1 && arr[i + 1] != 1) {
						continue;
					}
				if (i == n - 1) {
					if (arr[i] == 1) {
						ans += 1;
					}
				}
			}
			sb.append(ans + "\n");
		}
		System.out.println(sb);
	}
}
