#include <bits/stdc++.h>
using namespace std;
template <typename... T>
void show(T&&... args) {
  ((cerr << args << " "), ...);
}
template <typename F, typename S>
ostream& operator<<(ostream& os, const pair<F, S>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  typename vector<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[";
  typename set<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << "[";
  typename multiset<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename F, typename S>
ostream& operator<<(ostream& os, const map<F, S>& v) {
  os << "[";
  typename map<F, S>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
template <typename F, typename S>
ostream& operator<<(ostream& os, const unordered_map<F, S>& v) {
  os << "[";
  typename unordered_map<F, S>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
template <typename... T>
void read(T&... args) {
  ((cin >> args), ...);
}
template <typename... T>
void write(T&&... args) {
  ((cout << args << " "), ...);
}
int solve() {
  int n;
  string str[2];
  cin >> n >> str[0] >> str[1];
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int cnt = 0;
    if (str[0][i] == '0') cnt++;
    if (str[1][i] == '0') cnt++;
    if (cnt == 2) {
      if (i + 1 < n && str[0][i + 1] == '1' && str[1][i + 1] == '1') {
        i++;
        ans += 2;
      } else
        ans++;
    } else if (cnt == 0) {
      if (i + 1 < n && str[0][i + 1] == '0' && str[1][i + 1] == '0') {
        i++;
        ans += 2;
      }
    } else
      ans += 2;
  }
  cout << ans << "\n";
  return 0;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  ;
  ;
  int t = 1, cs = 0;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
void show(T&&... args) {
  ((cerr << args << " "), ...);
}
template <typename F, typename S>
ostream& operator<<(ostream& os, const pair<F, S>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  typename vector<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[";
  typename set<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << "[";
  typename multiset<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename F, typename S>
ostream& operator<<(ostream& os, const map<F, S>& v) {
  os << "[";
  typename map<F, S>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
template <typename F, typename S>
ostream& operator<<(ostream& os, const unordered_map<F, S>& v) {
  os << "[";
  typename unordered_map<F, S>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
template <typename... T>
void read(T&... args) {
  ((cin >> args), ...);
}
template <typename... T>
void write(T&&... args) {
  ((cout << args << " "), ...);
}
int solve() {
  int n;
  string str[2];
  cin >> n >> str[0] >> str[1];
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int cnt = 0;
    if (str[0][i] == '0') cnt++;
    if (str[1][i] == '0') cnt++;
    if (cnt == 2) {
      if (i + 1 < n && str[0][i + 1] == '1' && str[1][i + 1] == '1') {
        i++;
        ans += 2;
      } else
        ans++;
    } else if (cnt == 0) {
      if (i + 1 < n && str[0][i + 1] == '0' && str[1][i + 1] == '0') {
        i++;
        ans += 2;
      }
    } else
      ans += 2;
  }
  cout << ans << "\n";
  return 0;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  ;
  ;
  int t = 1, cs = 0;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```