#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
template <typename T>
using vc = vector<T>;
template <typename T>
using vvc = vc<vc<T>>;
template <typename T>
void smax(T& a, T b) {
  a = max(a, b);
}
template <typename T>
void smin(T& a, T b) {
  a = min(a, b);
}
void solve() {
  int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  bool mark[n];
  fill(mark, mark + n, false);
  int brojac = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] != t[i])
      brojac += 2;
    else if (s[i] == '0' && t[i] == '0') {
      brojac++;
      if (i && s[i - 1] == '1' && t[i - 1] == '1' && !mark[i - 1])
        brojac++;
      else if (i < n - 1 && s[i + 1] == '1' && t[i + 1] == '1') {
        brojac++;
        mark[i + 1] = true;
      }
    }
  }
  cout << brojac << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tt = 1;
  cin >> tt;
  while (tt--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
using ll = long long;
using ld = long double;
template <typename T>
using vc = vector<T>;
template <typename T>
using vvc = vc<vc<T>>;
template <typename T>
void smax(T& a, T b) {
  a = max(a, b);
}
template <typename T>
void smin(T& a, T b) {
  a = min(a, b);
}
void solve() {
  int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  bool mark[n];
  fill(mark, mark + n, false);
  int brojac = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] != t[i])
      brojac += 2;
    else if (s[i] == '0' && t[i] == '0') {
      brojac++;
      if (i && s[i - 1] == '1' && t[i - 1] == '1' && !mark[i - 1])
        brojac++;
      else if (i < n - 1 && s[i + 1] == '1' && t[i + 1] == '1') {
        brojac++;
        mark[i + 1] = true;
      }
    }
  }
  cout << brojac << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tt = 1;
  cin >> tt;
  while (tt--) solve();
  return 0;
}

```