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;

public class C_MAX_MEX_Cut {
  private static Scanner sc = new Scanner(System.in);

  public static void main(String[] args) {
    int t = sc.nextInt();
    while (t-- > 0) {
      solve();
    }
  }

  private static void solve() {
    int n = sc.nextInt();
    sc.nextLine();
    String s1 = sc.nextLine();
    String s2 = sc.nextLine();
    StringBuilder sb = new StringBuilder();
    int count = 0;
    for (int i = 0; i < s1.length(); i++) {
      if (s1.charAt(i) == '0' && s2.charAt(i) == '0') {
        sb.append(s1.charAt(i));
      } else if (s1.charAt(i) == '1' && s2.charAt(i) == '1') {
        sb.append(s1.charAt(i));
      } else {
        count += 2;
      }
    }
    if (sb.length() > 0) {
      char[] sArr = sb.toString().toCharArray();
      int i = 0;
      int j = 1;
      while (i < sArr.length && j < sArr.length) {
        if (sArr[i] == '1' && sArr[j] == '1') {
          j++;
        } else {
          if (sArr[i] == '0' && sArr[j] == '0') {
            count += 2;
          }
          if (sArr[i] == '0' && sArr[j] == '1') {
            count += 2;
          }
          if (sArr[i] == '1' && sArr[j] == '0') {
            count += 2;
          }
          j++;
          i = j;
          j++;
        }
      }
    }
    System.out.println(count);
  }
}
