#include <bits/stdc++.h>
using namespace std;
void precision(int a) { cout << setprecision(a) << fixed; }
vector<int> sieve(int n) {
  int* arr = new int[n + 1]();
  vector<int> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (long long j = (long long(i) * long long(i)); j <= n; j += i)
        arr[j] = 1;
    }
  return vect;
}
static bool comp(const pair<long long, long long>& a,
                 const pair<long long, long long>& b) {
  return a.second < b.second;
}
const long long N = 3e5 + 7;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<string> v;
    string s1, s2;
    cin >> s1 >> s2;
    v.push_back(s1);
    v.push_back(s2);
    long long cnt0 = 0, cnt1 = 0;
    for (long long i = 0; i < (2); i++) {
      for (long long j = 0; j < (n); j++) {
        if (v[i][j] == '0')
          cnt0++;
        else
          cnt1++;
      }
    }
    if (cnt0 == 0)
      cout << 0 << "\n";
    else if (cnt1 == 0)
      cout << n << "\n";
    else {
      long long cnt2 = 0, cnt3 = 0, cnt4 = 0, cnt5 = 0;
      for (long long i = 0; i < (n); i++) {
        if ((v[0][i] == '0' && v[1][i] == '1') ||
            (v[0][i] == '1' && v[1][i] == '0'))
          cnt2++;
        else if (v[0][i] == '0' && v[1][i] == '0') {
          if (i < n - 1 && v[0][i + 1] == '1' && v[1][i + 1] == '1') {
            cnt3++;
            i++;
          } else
            cnt4++;
        } else if (v[0][i] == '1' && v[1][i] == '1') {
          if (i < n - 1 && v[0][i + 1] == '0' && v[1][i + 1] == '0') {
            cnt5++;
            i++;
          }
        }
      }
      cout << 2 * cnt2 + 2 * cnt3 + cnt4 + 2 * cnt5 << "\n";
    }
  }
  return 0;
}

### Prompt

Generate 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;
void precision(int a) { cout << setprecision(a) << fixed; }
vector<int> sieve(int n) {
  int* arr = new int[n + 1]();
  vector<int> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (long long j = (long long(i) * long long(i)); j <= n; j += i)
        arr[j] = 1;
    }
  return vect;
}
static bool comp(const pair<long long, long long>& a,
                 const pair<long long, long long>& b) {
  return a.second < b.second;
}
const long long N = 3e5 + 7;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<string> v;
    string s1, s2;
    cin >> s1 >> s2;
    v.push_back(s1);
    v.push_back(s2);
    long long cnt0 = 0, cnt1 = 0;
    for (long long i = 0; i < (2); i++) {
      for (long long j = 0; j < (n); j++) {
        if (v[i][j] == '0')
          cnt0++;
        else
          cnt1++;
      }
    }
    if (cnt0 == 0)
      cout << 0 << "\n";
    else if (cnt1 == 0)
      cout << n << "\n";
    else {
      long long cnt2 = 0, cnt3 = 0, cnt4 = 0, cnt5 = 0;
      for (long long i = 0; i < (n); i++) {
        if ((v[0][i] == '0' && v[1][i] == '1') ||
            (v[0][i] == '1' && v[1][i] == '0'))
          cnt2++;
        else if (v[0][i] == '0' && v[1][i] == '0') {
          if (i < n - 1 && v[0][i + 1] == '1' && v[1][i + 1] == '1') {
            cnt3++;
            i++;
          } else
            cnt4++;
        } else if (v[0][i] == '1' && v[1][i] == '1') {
          if (i < n - 1 && v[0][i + 1] == '0' && v[1][i + 1] == '0') {
            cnt5++;
            i++;
          }
        }
      }
      cout << 2 * cnt2 + 2 * cnt3 + cnt4 + 2 * cnt5 << "\n";
    }
  }
  return 0;
}

```