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.Scanner;
import java.util.stream.IntStream;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);

    int t = sc.nextInt();
    for (int tc = 0; tc < t; ++tc) {
      sc.nextInt();
      String row1 = sc.next();
      String row2 = sc.next();

      System.out.println(solve(row1, row2));
    }

    sc.close();
  }

  static int solve(String row1, String row2) {
    int[] sums =
        IntStream.range(0, row1.length())
            .map(i -> (row1.charAt(i) - '0') + (row2.charAt(i) - '0'))
            .toArray();

    int result = 0;
    int index = 0;
    while (index != sums.length) {
      if (sums[index] == 0) {
        if (index != sums.length - 1 && sums[index + 1] == 2) {
          result += 2;
          index += 2;
        } else {
          ++result;
          ++index;
        }
      } else if (sums[index] == 1) {
        result += 2;
        ++index;
      } else {
        if (index != sums.length - 1 && sums[index + 1] == 0) {
          result += 2;
          index += 2;
        } else {
          ++index;
        }
      }
    }

    return result;
  }
}
