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.InputStreamReader;
import java.util.Arrays;

public class task_c {
	public static void main(String[] args) {
		try(BufferedReader reader = new BufferedReader(new InputStreamReader(System.in))) {
			StringBuilder sb = new StringBuilder();

			int count = Integer.parseInt(reader.readLine());
			for(int ii=0; ii<count; ii++) {
				int columns = Integer.parseInt(reader.readLine());
				int[] binary1 = Arrays.stream(reader.readLine().split("")).mapToInt(Integer::parseInt).toArray();
				int[] binary2 = Arrays.stream(reader.readLine().split("")).mapToInt(Integer::parseInt).toArray();
				int maxSum = 0;
				int reserved = -1;
				for(int i=0; i<columns; i++) {
					if(binary1[i]==binary2[i]) {
						if(binary1[i]==1) {
							if (i != columns - 1 && (binary1[i + 1] == 0 || binary2[i + 1] == 0)) {
								maxSum += 2;
								reserved = i + 1;
							} else if (i - 1 != reserved && i > 0 && (binary1[i - 1] == 0 || binary2[i - 1] == 0)) {
								maxSum++;
							}
						} else {
							if(i!=reserved)
								maxSum++;
						}
					} else if(i!=reserved) maxSum+=2;
				}
				sb.append(maxSum).append("\n");
			}

			System.out.println(sb);
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
}
