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.
#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long int n;
  cin >> n;
  string s1, s2;
  cin >> s1;
  cin >> s2;
  if (count(s1.begin(), s1.end(), '0') == n &&
      count(s2.begin(), s2.end(), '0') == n) {
    cout << n << "\n";
    return;
  } else if (count(s1.begin(), s1.end(), '1') == n &&
             count(s2.begin(), s2.end(), '1') == n) {
    cout << 0 << "\n";
    return;
  } else if (count(s1.begin(), s1.end(), '1') == n &&
             count(s2.begin(), s2.end(), '0') == n) {
    cout << 2 * n << "\n";
    return;
  } else if (count(s1.begin(), s1.end(), '0') == n &&
             count(s2.begin(), s2.end(), '1') == n) {
    cout << 2 * n << "\n";
    return;
  }
  long long int mex = 0;
  string aa = "";
  vector<string> a;
  for (long long int i = 0; i < n; i++) {
    if (s1.at(i) != s2.at(i)) {
      mex += 2;
      if (aa != "" || count(aa.begin(), aa.end(), '1') != aa.size()) {
        a.push_back(aa);
        aa = "";
      }
    } else {
      aa = aa + s1.at(i);
    }
  }
  if (aa != "" || count(aa.begin(), aa.end(), '1') != aa.size()) {
    a.push_back(aa);
    aa = "";
  }
  vector<string>::iterator i;
  for (i = a.begin(); i != a.end(); i++) {
    string s = *i;
    if (count(s.begin(), s.end(), '0') == s.size()) {
      mex += s.size();
    } else {
      for (long long int i = 0; i < s.length() - 1; i++) {
        if ((s.at(i) == '0' && s.at(i + 1) == '1') ||
            (s.at(i) == '1' && s.at(i + 1) == '0')) {
          s.at(i) = '*';
          s.at(i + 1) = '*';
          mex += 2;
        }
      }
      mex += count(s.begin(), s.end(), '0');
    }
  }
  cout << mex << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
