#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 200;
template <class Ruby_Rose>
inline void read(Ruby_Rose &x) {
  x = 0;
  char ch = getchar(), w = 0;
  while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = w ? -x : x;
}
int T, n;
char s[2][N];
int main() {
  read(T);
  while (T--) {
    read(n);
    scanf("%s%s", s[0] + 1, s[1] + 1);
    int las = -1, ans = 0;
    for (register int i = 1; i <= n; i++) {
      int t = s[0][i] - '0' + s[1][i] - '0';
      if (t == 0) {
        if (las == -1)
          las = 0;
        else if (las == 0)
          ans++;
        else if (las == 1)
          las = -1, ans += 2;
      } else if (t == 1) {
        if (las == 0) ans++;
        ans += 2;
        las = -1;
      } else if (t == 2) {
        if (las == -1)
          las = 1;
        else if (las == 0)
          las = -1, ans += 2;
      }
    }
    printf("%d\n", ans + (las == 0));
  }
  return 0;
}

### Prompt

Generate a Cpp solution 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 N = 1e6 + 200;
template <class Ruby_Rose>
inline void read(Ruby_Rose &x) {
  x = 0;
  char ch = getchar(), w = 0;
  while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = w ? -x : x;
}
int T, n;
char s[2][N];
int main() {
  read(T);
  while (T--) {
    read(n);
    scanf("%s%s", s[0] + 1, s[1] + 1);
    int las = -1, ans = 0;
    for (register int i = 1; i <= n; i++) {
      int t = s[0][i] - '0' + s[1][i] - '0';
      if (t == 0) {
        if (las == -1)
          las = 0;
        else if (las == 0)
          ans++;
        else if (las == 1)
          las = -1, ans += 2;
      } else if (t == 1) {
        if (las == 0) ans++;
        ans += 2;
        las = -1;
      } else if (t == 2) {
        if (las == -1)
          las = 1;
        else if (las == 0)
          las = -1, ans += 2;
      }
    }
    printf("%d\n", ans + (las == 0));
  }
  return 0;
}

```