#include <bits/stdc++.h>
const long long INF = 1000000001;
using namespace std;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t1;
  cin >> t1;
  for (long long q11 = 0; q11 < t1; q11++) {
    long long n;
    string s, t;
    cin >> n >> s >> t;
    vector<string> a = {""};
    long long ans = 0;
    for (long long q = 0; q < n; q++) {
      if (s[q] != t[q]) {
        ans += 2;
        if (!a.back().empty()) {
          a.push_back("");
        }
      } else {
        a.back().push_back(s[q]);
      }
    }
    for (string &f : a) {
      char last = '2';
      for (char q : f) {
        if (last == '2') {
          last = q;
        } else if (last == q) {
          ans += (q == '0');
        } else {
          ans += 2, last = '2';
        }
      }
      ans += (last == '0');
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
const long long INF = 1000000001;
using namespace std;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t1;
  cin >> t1;
  for (long long q11 = 0; q11 < t1; q11++) {
    long long n;
    string s, t;
    cin >> n >> s >> t;
    vector<string> a = {""};
    long long ans = 0;
    for (long long q = 0; q < n; q++) {
      if (s[q] != t[q]) {
        ans += 2;
        if (!a.back().empty()) {
          a.push_back("");
        }
      } else {
        a.back().push_back(s[q]);
      }
    }
    for (string &f : a) {
      char last = '2';
      for (char q : f) {
        if (last == '2') {
          last = q;
        } else if (last == q) {
          ans += (q == '0');
        } else {
          ans += 2, last = '2';
        }
      }
      ans += (last == '0');
    }
    cout << ans << '\n';
  }
  return 0;
}

```