#include <bits/stdc++.h>
using namespace std;
void KruNULL() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long prime(long long n) {
  long long flag = 1;
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      flag = 0;
      break;
    }
  }
  return flag;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  long long dono = 0;
  map<char, long long> m;
  long long zz = 0;
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    m[s[i]]++;
    m[t[i]]++;
    if (s[i] == '0' && t[i] == '0') zz++;
    if (s[i] != t[i]) {
      ans += (2ll + zz);
      zz = 0;
      m.clear();
    } else if ((long long)m.size() == 2) {
      ans += max({2ll, zz, 2ll + zz - 1});
      zz = 0;
      m.clear();
    }
  }
  ans += zz;
  cout << ans << endl;
}
int32_t main() {
  KruNULL();
  long long tc = 1;
  cin >> tc;
  for (long long i = 1; i <= tc; i++) {
    solve();
  }
}

### 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;
void KruNULL() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long prime(long long n) {
  long long flag = 1;
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      flag = 0;
      break;
    }
  }
  return flag;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  long long dono = 0;
  map<char, long long> m;
  long long zz = 0;
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    m[s[i]]++;
    m[t[i]]++;
    if (s[i] == '0' && t[i] == '0') zz++;
    if (s[i] != t[i]) {
      ans += (2ll + zz);
      zz = 0;
      m.clear();
    } else if ((long long)m.size() == 2) {
      ans += max({2ll, zz, 2ll + zz - 1});
      zz = 0;
      m.clear();
    }
  }
  ans += zz;
  cout << ans << endl;
}
int32_t main() {
  KruNULL();
  long long tc = 1;
  cin >> tc;
  for (long long i = 1; i <= tc; i++) {
    solve();
  }
}

```