#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
const int inf = 0x3f3f3f3f;
char s[2][maxn];
int ss[maxn];
int flag[maxn];
int n;
int main() {
  int _;
  scanf("%d", &_);
  while (_--) {
    scanf("%d", &n);
    scanf("%s", s[0] + 1);
    scanf("%s", s[1] + 1);
    memset(flag, 0, (n + 10) * sizeof(int));
    int num = 0;
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      if (s[0][i] == '0' && s[1][i] == '0') {
        num++;
        ss[i] = 0;
      }
      if (s[0][i] != s[1][i]) {
        ans += 2;
        ss[i] = 2;
      }
      if (s[0][i] == '1' && s[1][i] == '1') ss[i] = 1;
    }
    int cnt = 0;
    for (int i = 2; i <= n; i++) {
      if (((ss[i] == 1 && ss[i - 1] == 0) || (ss[i] == 0 && ss[i - 1] == 1)) &&
          flag[i - 1] == 0) {
        cnt++;
        flag[i] = flag[i - 1] = 1;
      }
    }
    printf("%d\n", num + cnt + ans);
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 maxn = 1e6 + 10;
const int inf = 0x3f3f3f3f;
char s[2][maxn];
int ss[maxn];
int flag[maxn];
int n;
int main() {
  int _;
  scanf("%d", &_);
  while (_--) {
    scanf("%d", &n);
    scanf("%s", s[0] + 1);
    scanf("%s", s[1] + 1);
    memset(flag, 0, (n + 10) * sizeof(int));
    int num = 0;
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      if (s[0][i] == '0' && s[1][i] == '0') {
        num++;
        ss[i] = 0;
      }
      if (s[0][i] != s[1][i]) {
        ans += 2;
        ss[i] = 2;
      }
      if (s[0][i] == '1' && s[1][i] == '1') ss[i] = 1;
    }
    int cnt = 0;
    for (int i = 2; i <= n; i++) {
      if (((ss[i] == 1 && ss[i - 1] == 0) || (ss[i] == 0 && ss[i - 1] == 1)) &&
          flag[i - 1] == 0) {
        cnt++;
        flag[i] = flag[i - 1] = 1;
      }
    }
    printf("%d\n", num + cnt + ans);
  }
  return 0;
}

```