#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
long long mul(long long a, long long b) { return (a * b) % (1000000007); }
long long add(long long a, long long b) { return (a + b) % (1000000007); }
long long sub(long long a, long long b) {
  return ((a - b) % (1000000007) + (1000000007)) % (1000000007);
}
void upd(long long &a, long long b) {
  a = (a % (1000000007) + b % (1000000007)) % (1000000007);
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
string s, s2;
int n;
int mex(int l, int r) {
  bool fl0 = 0, fl1 = 0;
  for (int i = l; i <= r; i++) {
    if (s[i] == '0')
      fl0 = 1;
    else
      fl1 = 1;
    if (s2[i] == '0')
      fl0 = 1;
    else
      fl1 = 1;
  }
  if (!fl0)
    return 0;
  else if (!fl1)
    return 1;
  else
    return 2;
}
int f[512345];
int main() {
  int T = read();
  while (T--) {
    n = read();
    cin >> s >> s2;
    int n = s.length();
    f[0] = mex(0, 0);
    for (int i = 1; i <= n - 1; i++) {
      f[i] = f[i - 1] + mex(i, i);
      if (i > 1) f[i] = max(f[i], f[i - 2] + mex(i - 1, i));
      if (i == 1) f[i] = max(f[i], mex(i - 1, i));
    }
    cout << f[n - 1] << endl;
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
#pragma comment(linker, "/STACK:102400000,102400000")
long long mul(long long a, long long b) { return (a * b) % (1000000007); }
long long add(long long a, long long b) { return (a + b) % (1000000007); }
long long sub(long long a, long long b) {
  return ((a - b) % (1000000007) + (1000000007)) % (1000000007);
}
void upd(long long &a, long long b) {
  a = (a % (1000000007) + b % (1000000007)) % (1000000007);
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
string s, s2;
int n;
int mex(int l, int r) {
  bool fl0 = 0, fl1 = 0;
  for (int i = l; i <= r; i++) {
    if (s[i] == '0')
      fl0 = 1;
    else
      fl1 = 1;
    if (s2[i] == '0')
      fl0 = 1;
    else
      fl1 = 1;
  }
  if (!fl0)
    return 0;
  else if (!fl1)
    return 1;
  else
    return 2;
}
int f[512345];
int main() {
  int T = read();
  while (T--) {
    n = read();
    cin >> s >> s2;
    int n = s.length();
    f[0] = mex(0, 0);
    for (int i = 1; i <= n - 1; i++) {
      f[i] = f[i - 1] + mex(i, i);
      if (i > 1) f[i] = max(f[i], f[i - 2] + mex(i - 1, i));
      if (i == 1) f[i] = max(f[i], mex(i - 1, i));
    }
    cout << f[n - 1] << endl;
  }
  return 0;
}

```