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 long long MOD = 1E9 + 7, INF = 2E18 + 5;
const double PI = 2 * acos(0.0);
const long double EPS = 1.0E-14;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long ans = 0, ans2 = 0, len;
    string s1, s2;
    cin >> len >> s1 >> s2;
    for (int i = 0; i < len; i++) {
      if ((s1[i] == '0' and s2[i] == '1') or (s1[i] == '1' and s2[i] == '0')) {
        ans += 2;
      } else if (s1[i] == '0' and s2[i] == '0') {
        ans++;
      } else if (s1[i] == '1' and s2[i] == '1') {
        if (i + 1 < len) {
          if (s1[i + 1] == '0' and s2[i + 1] == '0') {
            ans += 2;
            i++;
          }
        }
      }
    }
    reverse(s1.begin(), s1.end());
    reverse(s2.begin(), s2.end());
    for (int i = 0; i < len; i++) {
      if ((s1[i] == '0' and s2[i] == '1') or (s1[i] == '1' and s2[i] == '0')) {
        ans2 += 2;
      } else if (s1[i] == '0' and s2[i] == '0') {
        ans2++;
      } else if (s1[i] == '1' and s2[i] == '1') {
        if (i + 1 < len) {
          if (s1[i + 1] == '0' and s2[i + 1] == '0') {
            ans2 += 2;
            i++;
          }
        }
      }
    }
    cout << max(ans, ans2) << "\n";
  }
  return 0;
}
