#include <bits/stdc++.h>
using namespace std;
int t;
int n;
char a[100002];
char b[100002];
int ans;
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%s%s", &n, a + 1, b + 1);
    ans = 0;
    for (int i = 1; i <= n; ++i) {
      if (a[i] != b[i]) {
        ans += 2;
        continue;
      }
      if (a[i] == '1') {
        while (i <= n) {
          ++i;
          if (a[i] == '0' || b[i] == '0') {
            ans += 2;
            break;
          }
        }
      } else {
        if (i == n)
          ++ans;
        else if (a[i + 1] == '0' && b[i + 1] == '0')
          ++ans;
        else {
          if (a[i + 1] == '1' && b[i + 1] == '1')
            ans += 2;
          else
            ans += 3;
          ++i;
        }
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int t;
int n;
char a[100002];
char b[100002];
int ans;
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%s%s", &n, a + 1, b + 1);
    ans = 0;
    for (int i = 1; i <= n; ++i) {
      if (a[i] != b[i]) {
        ans += 2;
        continue;
      }
      if (a[i] == '1') {
        while (i <= n) {
          ++i;
          if (a[i] == '0' || b[i] == '0') {
            ans += 2;
            break;
          }
        }
      } else {
        if (i == n)
          ++ans;
        else if (a[i + 1] == '0' && b[i + 1] == '0')
          ++ans;
        else {
          if (a[i + 1] == '1' && b[i + 1] == '1')
            ans += 2;
          else
            ans += 3;
          ++i;
        }
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

```