#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream &operator<<(ostream &output, const vector<T> &v) {
  if (v.empty()) return output;
  for (int i = 0; i < v.size() - 1; i++) output << v[i] << " ";
  output << v.back();
  return output;
}
template <typename T>
istream &operator>>(istream &input, vector<T> &v) {
  for (auto &i : v) cin >> i;
  return input;
}
void __sol() {
  int n;
  cin >> n;
  string s[2];
  for (int i = 0; i < 2; i++) cin >> s[i];
  for (int i = 0; i < n; i++) {
    if (s[0][i] > s[1][i]) swap(s[0][i], s[1][i]);
  }
  vector<int> vis(n + 2, 0);
  auto get = [&](int p, int q) {
    int ans = 0;
    for (int i = p; i <= q; i++) {
      if (!vis[i]) {
        if (i - 1 >= 0 && !vis[i - 1] && s[0][i - 1] != s[0][i]) {
          vis[i] = 1;
          ans += 2;
        } else if (i + 1 <= q && s[0][i + 1] != s[0][i]) {
          vis[i] = 1;
          vis[i + 1] = 1;
          ans += 2;
        } else if (s[0][i] == '0')
          ans++, vis[i] = 1;
      }
    }
    return ans;
  };
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int j = i;
    while (j < n && !(s[0][j] == '0' && s[1][j] == '1')) {
      j++;
    }
    ans += get(i, j - 1);
    if (j < n) ans += 2, vis[j] = 1;
    i = j;
  }
  cout << ans << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) __sol();
  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;
template <typename T>
ostream &operator<<(ostream &output, const vector<T> &v) {
  if (v.empty()) return output;
  for (int i = 0; i < v.size() - 1; i++) output << v[i] << " ";
  output << v.back();
  return output;
}
template <typename T>
istream &operator>>(istream &input, vector<T> &v) {
  for (auto &i : v) cin >> i;
  return input;
}
void __sol() {
  int n;
  cin >> n;
  string s[2];
  for (int i = 0; i < 2; i++) cin >> s[i];
  for (int i = 0; i < n; i++) {
    if (s[0][i] > s[1][i]) swap(s[0][i], s[1][i]);
  }
  vector<int> vis(n + 2, 0);
  auto get = [&](int p, int q) {
    int ans = 0;
    for (int i = p; i <= q; i++) {
      if (!vis[i]) {
        if (i - 1 >= 0 && !vis[i - 1] && s[0][i - 1] != s[0][i]) {
          vis[i] = 1;
          ans += 2;
        } else if (i + 1 <= q && s[0][i + 1] != s[0][i]) {
          vis[i] = 1;
          vis[i + 1] = 1;
          ans += 2;
        } else if (s[0][i] == '0')
          ans++, vis[i] = 1;
      }
    }
    return ans;
  };
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int j = i;
    while (j < n && !(s[0][j] == '0' && s[1][j] == '1')) {
      j++;
    }
    ans += get(i, j - 1);
    if (j < n) ans += 2, vis[j] = 1;
    i = j;
  }
  cout << ans << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) __sol();
  return 0;
}

```