#include <bits/stdc++.h>
using namespace std;
const int iinf = 1e9 + 7;
const long long int inf = 1e9 + 7;
const long long int mod = 998244353;
const long long int maxn = 2e5 + 5;
long long int sqr(long long int x) { return x * x; }
long long int modd(long long int a) {
  if (a > 0) return a % inf;
  while (a < 0) {
    a += inf;
  }
  return a;
}
long long int modexpo(long long int a, long long int b, long long int iinf1) {
  a %= iinf1;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % iinf1;
    a = a * a % iinf1;
    b >>= 1;
  }
  return res;
}
void solve() {
  int n;
  cin >> n;
  string s, s1;
  cin >> s >> s1;
  long long int ans = 0;
  vector<bool> vis(n, false);
  for (int i = 0; i < n - 1; i++) {
    if (s[i] != s1[i]) {
      ans += 2;
      vis[i] = true;
    } else {
      if (s[i + 1] == s1[i + 1] && s[i] != s[i + 1]) {
        vis[i] = true;
        vis[i + 1] = true;
        ans += 2;
        i++;
      } else {
        if (s[i] == '0') {
          ans += 1;
        }
        vis[i] = true;
      }
    }
  }
  if (!vis[n - 1]) {
    if (s[n - 1] != s1[n - 1]) {
      ans += 2;
    } else {
      if (s[n - 1] == '0') ans++;
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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;
const int iinf = 1e9 + 7;
const long long int inf = 1e9 + 7;
const long long int mod = 998244353;
const long long int maxn = 2e5 + 5;
long long int sqr(long long int x) { return x * x; }
long long int modd(long long int a) {
  if (a > 0) return a % inf;
  while (a < 0) {
    a += inf;
  }
  return a;
}
long long int modexpo(long long int a, long long int b, long long int iinf1) {
  a %= iinf1;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % iinf1;
    a = a * a % iinf1;
    b >>= 1;
  }
  return res;
}
void solve() {
  int n;
  cin >> n;
  string s, s1;
  cin >> s >> s1;
  long long int ans = 0;
  vector<bool> vis(n, false);
  for (int i = 0; i < n - 1; i++) {
    if (s[i] != s1[i]) {
      ans += 2;
      vis[i] = true;
    } else {
      if (s[i + 1] == s1[i + 1] && s[i] != s[i + 1]) {
        vis[i] = true;
        vis[i + 1] = true;
        ans += 2;
        i++;
      } else {
        if (s[i] == '0') {
          ans += 1;
        }
        vis[i] = true;
      }
    }
  }
  if (!vis[n - 1]) {
    if (s[n - 1] != s1[n - 1]) {
      ans += 2;
    } else {
      if (s[n - 1] == '0') ans++;
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```