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;
const int N = 1e5 + 10;
char a[2][N];
bool s[N];
void solve() {
  int n;
  cin >> n;
  a[0][0] = a[1][0] = '5';
  a[0][n + 1] = a[1][n + 1] = '5';
  memset(s, true, sizeof s);
  for (int i = 0; i < 2; i++) {
    for (int j = 1; j <= n; j++) cin >> a[i][j];
  }
  int res = 0;
  for (int i = 1; i <= n; i++) {
    if (a[0][i] == '0' && a[1][i] == '1' || a[0][i] == '1' && a[1][i] == '0') {
      res += 2;
    } else {
      if (a[0][i] == '0' && s[i] == true) {
        if (a[0][i - 1] == '1' && a[1][i - 1] == '1' && s[i - 1] == true) {
          res += 2;
          s[i] = false;
          s[i - 1] = false;
        } else if (a[0][i + 1] == '1' && a[1][i + 1] == '1' &&
                   s[i + 1] == true) {
          res += 2;
          s[i] = false;
          s[i + 1] = false;
        } else if (a[1][i - 1] == '0' && a[0][i - 1] == '0' &&
                   s[i - 1] == true) {
          s[i - 1] = false;
          res++;
        } else if (a[0][i + 1] == '0' && a[1][i + 1] == '1' ||
                   a[1][i + 1] == '1' && a[0][i + 1] == '0' && s[i] == true) {
          s[i] = false;
          res++;
        }
      } else if (a[1][i] == '1' && s[i] == true) {
        if (a[1][i - 1] == '0' && a[0][i - 1] == '0' && s[i - 1] == true) {
          res += 2;
          s[i - 1] = false;
          s[i] = false;
        } else if (a[0][i + 1] == '0' && a[1][i + 1] == '0' &&
                   s[i + 1] == true) {
          res += 2;
          s[i + 1] = false;
          s[i] = false;
        }
      }
    }
  }
  if (a[0][n] == '0' && a[1][n] == '0' && s[n] == true) res++;
  cout << res << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
