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 run() {
  int n;
  cin >> n;
  string s[2];
  cin >> s[0];
  cin >> s[1];
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (s[0][i] != s[1][i]) {
      ans += 2;
    } else if (s[0][i] == '0' && s[1][i] == '0') {
      ans += 1;
    }
  }
  vector<int> typ;
  for (int i = 0; i + 1 < n; i++) {
    if (s[0][i] == '0' && s[1][i] == '0' && s[0][i + 1] == '1' &&
        s[1][i + 1] == '1') {
      typ.push_back(i);
    } else if (s[0][i] == '1' && s[1][i] == '1' && s[0][i + 1] == '0' &&
               s[1][i + 1] == '0') {
      typ.push_back(i);
    }
  }
  int k = (int)typ.size();
  if (k == 0) {
    ;
  } else if (k == 1) {
    ans += 1;
  } else {
    vector<int> dp(k);
    dp[0] = 1;
    dp[1] = 1 + (typ[1] != typ[0] + 1);
    for (int i = 2; i < k; i++) {
      dp[i] = dp[i - 2] + 1;
      if (typ[i] != typ[i - 1] + 1) dp[i] = max(dp[i], dp[i - 1] + 1);
    }
    ans += dp.back();
  }
  cout << ans << '\n';
}
int main() {
  int t;
  cin >> t;
  while (t--) run();
}
