#include <bits/stdc++.h>
int main() {
  int t;
  scanf("%d ", &t);
  char *a, *b;
  for (int i = 0; i < t; i++) {
    int n;
    int max = 0;
    scanf("%d ", &n);
    a = (char *)malloc(sizeof(char) * n);
    b = (char *)malloc(sizeof(char) * n);
    for (int j = 0; j < n; j++) {
      if (j == n - 1) {
        scanf("%c ", &a[j]);
      } else {
        scanf("%c", &a[j]);
      }
    }
    for (int j = 0; j < n; j++) {
      scanf("%c", &b[j]);
    }
    int pre, cur;
    if (a[0] == '1') {
      if (b[0] == '1') {
        cur = 3;
      } else {
        cur = 2;
      }
    } else {
      if (b[0] == '1') {
        cur = 1;
      } else {
        cur = 0;
      }
    }
    for (int j = 1; j < n; j++) {
      pre = cur;
      if (a[j] == '1') {
        if (b[j] == '1') {
          cur = 3;
        } else {
          cur = 2;
        }
      } else {
        if (b[j] == '1') {
          cur = 1;
        } else {
          cur = 0;
        }
      }
      if ((pre == 1) || (pre == 2)) {
        max += 2;
      } else if (pre == 0) {
        if (cur == 3) {
          cur = 1;
        } else {
          max += 1;
        }
      } else {
        if (cur == 0) {
          cur = 1;
        }
      }
    }
    if ((cur == 1) || (cur == 2)) {
      max += 2;
    } else if (cur == 0) {
      max += 1;
    }
    printf("%d\n", max);
  }
  return 0;
}

### 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>
int main() {
  int t;
  scanf("%d ", &t);
  char *a, *b;
  for (int i = 0; i < t; i++) {
    int n;
    int max = 0;
    scanf("%d ", &n);
    a = (char *)malloc(sizeof(char) * n);
    b = (char *)malloc(sizeof(char) * n);
    for (int j = 0; j < n; j++) {
      if (j == n - 1) {
        scanf("%c ", &a[j]);
      } else {
        scanf("%c", &a[j]);
      }
    }
    for (int j = 0; j < n; j++) {
      scanf("%c", &b[j]);
    }
    int pre, cur;
    if (a[0] == '1') {
      if (b[0] == '1') {
        cur = 3;
      } else {
        cur = 2;
      }
    } else {
      if (b[0] == '1') {
        cur = 1;
      } else {
        cur = 0;
      }
    }
    for (int j = 1; j < n; j++) {
      pre = cur;
      if (a[j] == '1') {
        if (b[j] == '1') {
          cur = 3;
        } else {
          cur = 2;
        }
      } else {
        if (b[j] == '1') {
          cur = 1;
        } else {
          cur = 0;
        }
      }
      if ((pre == 1) || (pre == 2)) {
        max += 2;
      } else if (pre == 0) {
        if (cur == 3) {
          cur = 1;
        } else {
          max += 1;
        }
      } else {
        if (cur == 0) {
          cur = 1;
        }
      }
    }
    if ((cur == 1) || (cur == 2)) {
      max += 2;
    } else if (cur == 0) {
      max += 1;
    }
    printf("%d\n", max);
  }
  return 0;
}

```