#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  string f, s;
  cin >> f >> s;
  int one = 0, zero = 0, st = 0, ans = 0;
  for (int i = 0; i < n; ++i) {
    if (f[i] == '1')
      ++one;
    else
      ++zero;
    if (s[i] == '1')
      ++one;
    else
      ++zero;
    ++st;
    if (zero > 0) {
      if (zero > 1 && st > 1 && one == 1) {
        ans += 1;
        zero -= 2;
      }
      if (one > 0) {
        ans += 2, st = 0, one = 0, zero = 0;
      } else if (st > 1) {
        st = 1, ans += 1, zero -= 2;
      }
    }
  }
  if (zero > 0) {
    if (one > 0) {
      ans += 2, st = 0, one = 0, zero = 0;
    } else {
      ans += 1;
    }
  }
  cout << ans << "\n";
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please formulate 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;
void solve() {
  int n;
  cin >> n;
  string f, s;
  cin >> f >> s;
  int one = 0, zero = 0, st = 0, ans = 0;
  for (int i = 0; i < n; ++i) {
    if (f[i] == '1')
      ++one;
    else
      ++zero;
    if (s[i] == '1')
      ++one;
    else
      ++zero;
    ++st;
    if (zero > 0) {
      if (zero > 1 && st > 1 && one == 1) {
        ans += 1;
        zero -= 2;
      }
      if (one > 0) {
        ans += 2, st = 0, one = 0, zero = 0;
      } else if (st > 1) {
        st = 1, ans += 1, zero -= 2;
      }
    }
  }
  if (zero > 0) {
    if (one > 0) {
      ans += 2, st = 0, one = 0, zero = 0;
    } else {
      ans += 1;
    }
  }
  cout << ans << "\n";
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```