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, j = 0;
			int ans = 0;

			int i = 0;
			while (i < n) {
				if (s[i] == '1' && p[i] == '1') {
					if (cnt != 0) {
						ans += 2;
						cnt = 0;
					} else j = 1;
					++i;
					continue;
				}

				if (s[i] == '0' && p[i] == '0') {
					if (j != 0) {
						ans += 2;
						j = 0;
					} else if (cnt != 0) ans++;
					else cnt = 1;
					++i;
					continue;
				}
				if (cnt != 0) ans++;
				ans += 2;
				cnt = j = 0;
				++i;
			}
			if (cnt != 0) ++ans;
			out.println(ans);
		}
		out.close();
	}
	
	
}