#include <bits/stdc++.h>
using namespace std;
static long long int mod = (long long)(1e9) + 7;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
vector<int> prime((long long)(1e7), 1);
void seive(long long n) {
  long long i, j;
  prime[0] = prime[1] = 0;
  for (i = 2; i * i <= n; i++) {
    if (prime[i] == 1) {
      for (j = i * i; j <= n; j += i) prime[j] = 0;
    }
  }
}
void solve() {
  char ch[] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm',
               'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
  int n;
  cin >> n;
  string str1, str2;
  cin >> str1;
  cin >> str2;
  int i, ans = 0, zeros = 0;
  map<char, int> m1;
  for (i = 0; i < n - 1; i++) {
    map<char, int> m1;
    m1[str1[i]]++;
    m1[str2[i]]++;
    if (m1['0'] > 0 && m1['1'] > 0)
      ans += 2;
    else {
      m1.clear();
      m1[str1[i + 1]]++;
      m1[str2[i + 1]]++;
      if (str1[i] == '0' && str2[i] == '0' && m1['1'] == 2) {
        ans += 2;
        i++;
      } else if (str1[i] == '1' && str2[i] == '1' &&
                 (str1[i + 1] == '0' || str2[i + 1] == '0')) {
        ans += 2;
        i++;
      } else if (str1[i] == '0' && str2[i] == '0')
        ans++;
    }
  }
  if (i == n)
    ;
  else {
    map<char, int> m1;
    m1[str1[i]]++;
    m1[str2[i]]++;
    if (m1['0'] > 0 && m1['1'] > 0)
      ans += 2;
    else if (m1['0'] > 0)
      ans++;
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Create a solution in CPP for 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;
static long long int mod = (long long)(1e9) + 7;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
vector<int> prime((long long)(1e7), 1);
void seive(long long n) {
  long long i, j;
  prime[0] = prime[1] = 0;
  for (i = 2; i * i <= n; i++) {
    if (prime[i] == 1) {
      for (j = i * i; j <= n; j += i) prime[j] = 0;
    }
  }
}
void solve() {
  char ch[] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm',
               'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
  int n;
  cin >> n;
  string str1, str2;
  cin >> str1;
  cin >> str2;
  int i, ans = 0, zeros = 0;
  map<char, int> m1;
  for (i = 0; i < n - 1; i++) {
    map<char, int> m1;
    m1[str1[i]]++;
    m1[str2[i]]++;
    if (m1['0'] > 0 && m1['1'] > 0)
      ans += 2;
    else {
      m1.clear();
      m1[str1[i + 1]]++;
      m1[str2[i + 1]]++;
      if (str1[i] == '0' && str2[i] == '0' && m1['1'] == 2) {
        ans += 2;
        i++;
      } else if (str1[i] == '1' && str2[i] == '1' &&
                 (str1[i + 1] == '0' || str2[i + 1] == '0')) {
        ans += 2;
        i++;
      } else if (str1[i] == '0' && str2[i] == '0')
        ans++;
    }
  }
  if (i == n)
    ;
  else {
    map<char, int> m1;
    m1[str1[i]]++;
    m1[str2[i]]++;
    if (m1['0'] > 0 && m1['1'] > 0)
      ans += 2;
    else if (m1['0'] > 0)
      ans++;
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```