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;
int toD(string b) {
  int n = 0;
  for (int i = 0; i < b.size(); ++i) {
    if (b[b.size() - 1 - i] == '1') n += pow(2, i);
  }
  return n;
}
void solve() {
  int n;
  cin >> n;
  string bit1, bit2;
  cin >> bit1 >> bit2;
  int a, b;
  a = toD(bit1);
  b = toD(bit2);
  vector<int> r;
  for (int i = 1; i <= n; ++i) {
    if ((a) & (1 << (n - i)) && (b) & (1 << (n - i)))
      r.push_back(0);
    else if (!((a) & (1 << (n - i))) && !((b) & (1 << (n - i))))
      r.push_back(1);
    else
      r.push_back(2);
  }
  int sum = 0;
  for (int i = 0; i < r.size(); i++) {
    if (r[i] == 2)
      sum += 2;
    else {
      if (i == r.size() - 1)
        sum += r[i];
      else if (r[i] != r[i + 1]) {
        sum += 2;
        i++;
      } else
        sum += r[i];
    }
  }
  cout << sum << '\n';
}
int main() {
  int n;
  cin >> n;
  while (n--) {
    solve();
  }
}
