#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b);
bool prime(int n);
long long sumdigit(long long n);
bool ispowerof2(int n);
long long onesinbinary(long long n);
long long power(long long a, long long b);
bool cmp(pair<int, int> x, pair<int, int> y) { return x.second < y.second; }
bool cmp(int x, int y) { return x > y; }
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    char arr[2][n];
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < n; j++) {
        cin >> arr[i][j];
      }
    }
    int ans = 0;
    vector<int> v;
    for (int i = 0; i < n; i++) {
      if (arr[0][i] == '0' && arr[1][i] == '0')
        ans++;
      else if (arr[0][i] == '0' || arr[1][i] == '0')
        ans += 2;
      else {
        v.push_back(i);
      }
    }
    int x = v.size();
    for (int i = 0; i < x; i++) {
      if (v[i] < n - 1 && v[i] > 0) {
        if (arr[0][v[i] - 1] == '0' && arr[1][v[i] - 1] == '0') {
          ans++;
        } else if (arr[0][v[i] + 1] == '0' && arr[1][v[i] + 1] == '0') {
          ans++;
          arr[0][v[i] + 1] = '1', arr[1][v[i] + 1] = '1';
        }
      } else if (v[i] == n - 1) {
        if (arr[0][v[i] - 1] == '0' && arr[1][v[i] - 1] == '0') {
          ans++;
        }
      } else if (v[i] == 0) {
        if (arr[0][v[i] + 1] == '0' && arr[1][v[i] + 1] == '0') {
          ans++;
          arr[0][v[i] + 1] = '1', arr[1][v[i] + 1] = '1';
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}
long long onesinbinary(long long n) {
  long long cnt = 0;
  while (n) {
    n = n & (n - 1);
    cnt++;
  }
  return cnt;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool prime(int n) {
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long sumdigit(long long n) {
  int c = 0;
  while (n > 0) {
    c += n % 10;
    n /= 10;
  }
  return c;
}
bool ispowerof2(int n) {
  if (n == 0) return false;
  return (!(n & ((n) & (n - 1))));
}
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long res = 1;
  while (b) {
    if (b % 2 == 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b /= 2;
  }
  return res;
}

### Prompt

Please create a solution in cpp 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;
int gcd(int a, int b);
bool prime(int n);
long long sumdigit(long long n);
bool ispowerof2(int n);
long long onesinbinary(long long n);
long long power(long long a, long long b);
bool cmp(pair<int, int> x, pair<int, int> y) { return x.second < y.second; }
bool cmp(int x, int y) { return x > y; }
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    char arr[2][n];
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < n; j++) {
        cin >> arr[i][j];
      }
    }
    int ans = 0;
    vector<int> v;
    for (int i = 0; i < n; i++) {
      if (arr[0][i] == '0' && arr[1][i] == '0')
        ans++;
      else if (arr[0][i] == '0' || arr[1][i] == '0')
        ans += 2;
      else {
        v.push_back(i);
      }
    }
    int x = v.size();
    for (int i = 0; i < x; i++) {
      if (v[i] < n - 1 && v[i] > 0) {
        if (arr[0][v[i] - 1] == '0' && arr[1][v[i] - 1] == '0') {
          ans++;
        } else if (arr[0][v[i] + 1] == '0' && arr[1][v[i] + 1] == '0') {
          ans++;
          arr[0][v[i] + 1] = '1', arr[1][v[i] + 1] = '1';
        }
      } else if (v[i] == n - 1) {
        if (arr[0][v[i] - 1] == '0' && arr[1][v[i] - 1] == '0') {
          ans++;
        }
      } else if (v[i] == 0) {
        if (arr[0][v[i] + 1] == '0' && arr[1][v[i] + 1] == '0') {
          ans++;
          arr[0][v[i] + 1] = '1', arr[1][v[i] + 1] = '1';
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}
long long onesinbinary(long long n) {
  long long cnt = 0;
  while (n) {
    n = n & (n - 1);
    cnt++;
  }
  return cnt;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool prime(int n) {
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long sumdigit(long long n) {
  int c = 0;
  while (n > 0) {
    c += n % 10;
    n /= 10;
  }
  return c;
}
bool ispowerof2(int n) {
  if (n == 0) return false;
  return (!(n & ((n) & (n - 1))));
}
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long res = 1;
  while (b) {
    if (b % 2 == 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b /= 2;
  }
  return res;
}

```