#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
long long t;
long long mex(vector<long long> &v) {
  sort(v.begin(), v.end());
  long long ans = 0;
  for (long long x : v)
    if (x == ans) ans++;
  return ans;
}
long long main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    vector<long long> dp(n);
    function<long long(long long, long long)> get_mex = [&](long long i,
                                                            long long delta) {
      vector<long long> v;
      for (long long j = 0; j < delta; j++) {
        v.push_back(s1[i - j] - '0');
        v.push_back(s2[i - j] - '0');
      }
      return mex(v);
    };
    for (long long i = 0; i < n; i++) {
      for (long long prv = i - 1; prv >= i - 2; prv--) {
        long long delta = i - prv;
        if (prv == -1) {
          dp[i] = max(dp[i], get_mex(i, delta));
        } else if (prv < -1) {
          continue;
        } else {
          dp[i] = max(dp[i], dp[prv] + get_mex(i, delta));
        }
      }
    }
    cout << dp[n - 1] << "\n";
  }
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
long long t;
long long mex(vector<long long> &v) {
  sort(v.begin(), v.end());
  long long ans = 0;
  for (long long x : v)
    if (x == ans) ans++;
  return ans;
}
long long main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    vector<long long> dp(n);
    function<long long(long long, long long)> get_mex = [&](long long i,
                                                            long long delta) {
      vector<long long> v;
      for (long long j = 0; j < delta; j++) {
        v.push_back(s1[i - j] - '0');
        v.push_back(s2[i - j] - '0');
      }
      return mex(v);
    };
    for (long long i = 0; i < n; i++) {
      for (long long prv = i - 1; prv >= i - 2; prv--) {
        long long delta = i - prv;
        if (prv == -1) {
          dp[i] = max(dp[i], get_mex(i, delta));
        } else if (prv < -1) {
          continue;
        } else {
          dp[i] = max(dp[i], dp[prv] + get_mex(i, delta));
        }
      }
    }
    cout << dp[n - 1] << "\n";
  }
}

```